149753
A wheel of bicycle is rolling without slipping on a level road. The velocity of the centre of mass is \(\mathbf{v}_{\mathrm{CM}}\), then true statement is
1 The velocity of point \(A\) is \(2 v_{\mathrm{CM}}\) and velocity of point \(B\) is zero
2 The velocity of point \(A\) is zero and velocity of point \(B\) is \(2 \mathrm{v}_{\mathrm{CM}}\)
3 The velocity of point \(A\) is \(2 \mathrm{v}_{\mathrm{CM}}\) and velocity of point \(B\) is \(-v_{C M}\)
4 The velocities of both \(A\) and \(B\) are \(v_{C M}\)
Explanation:
A Let the circle radius is \(\mathrm{R}\) and with \(\omega\). The velocity of centre of mass of body rolling without slipping is given by. \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\) velocity at point \(\mathrm{A}-\) \(\mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{cm}}+\mathrm{R} \omega\) \(=\mathrm{v}_{\mathrm{cm}}+\mathrm{v}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{A}}=2 \mathrm{v}_{\mathrm{cm}}\) velocity at point \(\mathrm{B}-\) \(\mathrm{v}_{\mathrm{B}}=\mathrm{v}_{\mathrm{cm}}-\mathrm{R} \omega\) \(=\mathrm{V}_{\mathrm{cm}}-\mathrm{V}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{B}}=0\)
AIPMT-2001
Rotational Motion
149754
A light rod of length \(l\) has two masses \(m_{1}\) and \(m_{2}\) attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
A To the given data, Length of \(\operatorname{rod}=l\) masses \(=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) From the above figure \(\mathrm{r}_{1} =\frac{\mathrm{m}_{1} \times 0+\mathrm{m}_{2} \times l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{r}_{2} =\frac{\mathrm{m}_{1} \times l+\mathrm{m}_{2} \times 0}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{1} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) Now at the moment of inertia about centre of mass \(\mathrm{I} =\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2}^{2} \mathrm{r}_{2}^{2}\) \(=\mathrm{m}_{1} \times \frac{\mathrm{m}_{2}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}+\mathrm{m}_{2} \times \frac{\mathrm{m}_{1}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\) \(=\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{I} =\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
NEET- 2016
Rotational Motion
149757
A rod is of length \(3 \mathrm{~m}\) and its mass acting per unit length is directly proportional to distance \(x\) from its one end. The centre of gravity of the rod from that end will be at
1 \(1.5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(2.5 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
B Assuming consider an elementary length \(\mathrm{dx}\) Mass of small element of \(\mathrm{dx}\) length is, \(\mathrm{dm}=\mathrm{kx} . \mathrm{dx}\) [where \(\mathrm{k}\) is the proportionality constant] At the centre of gravity of the \(\operatorname{rod}\left(x_{d}\right)\) is given. \(\mathrm{x}_{\mathrm{d}}=\frac{\int_{0}^{3} \mathrm{x} \cdot \mathrm{dm}}{\int_{0}^{3} \mathrm{dm}}\) \(\left(x_{d}\right)=\frac{\int_{0}^{3} k x \cdot d x \cdot x}{\int_{0}^{3} k x \cdot d x}=\frac{\int_{0}^{3} x^{2} \cdot d x}{\int_{0}^{3} x \cdot d x}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{3}}{\left[\frac{x^{2}}{2}\right]_{0}^{3}}\) Or \(\quad\left(\mathrm{x}_{\mathrm{d}}\right)=\frac{27 / 3}{9 / 2}=2 \mathrm{~m}\)
AIPMT-2002
Rotational Motion
149759
Two bodies of masses \(1 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) have position vectors \(\hat{i}+2 \hat{j}+\hat{k}\) and \(-3 \hat{i}-2 \hat{j}+\hat{k}\), respectively. The centre of mass of this system has a position vector
1 \(-2 \hat{i}+2 \hat{k}\)
2 \(-2 \hat{i}-\hat{j}+\hat{k}\)
3 \(2 \hat{i}-\hat{j}-2 \hat{k}\)
4 \(-\hat{i}+\hat{j}+\hat{k}\)
Explanation:
B Given that, Masses of bodies - \(\mathrm{M}_{1}=1 \mathrm{~kg} \& \mathrm{M}_{2}=3 \mathrm{~kg}\) Position vectors are - \(\overrightarrow{\mathrm{r}}_{1}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{r}}_{2}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) Position vector of the centre of mass \(\vec{r}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{2} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(= \frac{1 \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+3(-3 \hat{i}-2 \hat{j}+\hat{\mathrm{k}})}{1+3}\) \(= \frac{-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{4}=\frac{4(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})}{4}\) \(\overrightarrow{\mathrm{r}}_{\mathrm{cm}}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
AIPMT-2009
Rotational Motion
149761
A rod of weight \(w\) is supported by two parallel knife edges \(A\) and \(B\) and is in equilibrium in a horizontal position. The knives are at a distanced \(d\) from each other. The centre of mass of the rod is at distance \(x\) from \(A\). The normal reaction on \(A\) is
D According to question, situation is shown in below figure- Where, \(N_{A}=\text { Normal reaction on } A\) \(N_{B}=\text { Normal reaction on } B\) \(W=\text { weight of rod }\) In equilibrium position, \(\mathrm{N}_{\mathrm{A}}+\mathrm{N}_{\mathrm{B}}=\mathrm{w}\) By Torque balancing about centre of mass of the rod, \(\mathrm{N}_{\mathrm{A}} \mathrm{x}=\mathrm{N}_{\mathrm{B}}(\mathrm{d}-\mathrm{x})\) Putting the value of \(N_{B}\) from equation (i) \(N_{A} x=\left(w-N_{A}\right)(d-x)\) \(N_{A} x=w d-w x-N_{A} d+N_{A} x\) \(N_{A}=\frac{w(d-x)}{d}\)
149753
A wheel of bicycle is rolling without slipping on a level road. The velocity of the centre of mass is \(\mathbf{v}_{\mathrm{CM}}\), then true statement is
1 The velocity of point \(A\) is \(2 v_{\mathrm{CM}}\) and velocity of point \(B\) is zero
2 The velocity of point \(A\) is zero and velocity of point \(B\) is \(2 \mathrm{v}_{\mathrm{CM}}\)
3 The velocity of point \(A\) is \(2 \mathrm{v}_{\mathrm{CM}}\) and velocity of point \(B\) is \(-v_{C M}\)
4 The velocities of both \(A\) and \(B\) are \(v_{C M}\)
Explanation:
A Let the circle radius is \(\mathrm{R}\) and with \(\omega\). The velocity of centre of mass of body rolling without slipping is given by. \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\) velocity at point \(\mathrm{A}-\) \(\mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{cm}}+\mathrm{R} \omega\) \(=\mathrm{v}_{\mathrm{cm}}+\mathrm{v}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{A}}=2 \mathrm{v}_{\mathrm{cm}}\) velocity at point \(\mathrm{B}-\) \(\mathrm{v}_{\mathrm{B}}=\mathrm{v}_{\mathrm{cm}}-\mathrm{R} \omega\) \(=\mathrm{V}_{\mathrm{cm}}-\mathrm{V}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{B}}=0\)
AIPMT-2001
Rotational Motion
149754
A light rod of length \(l\) has two masses \(m_{1}\) and \(m_{2}\) attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
A To the given data, Length of \(\operatorname{rod}=l\) masses \(=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) From the above figure \(\mathrm{r}_{1} =\frac{\mathrm{m}_{1} \times 0+\mathrm{m}_{2} \times l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{r}_{2} =\frac{\mathrm{m}_{1} \times l+\mathrm{m}_{2} \times 0}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{1} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) Now at the moment of inertia about centre of mass \(\mathrm{I} =\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2}^{2} \mathrm{r}_{2}^{2}\) \(=\mathrm{m}_{1} \times \frac{\mathrm{m}_{2}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}+\mathrm{m}_{2} \times \frac{\mathrm{m}_{1}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\) \(=\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{I} =\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
NEET- 2016
Rotational Motion
149757
A rod is of length \(3 \mathrm{~m}\) and its mass acting per unit length is directly proportional to distance \(x\) from its one end. The centre of gravity of the rod from that end will be at
1 \(1.5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(2.5 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
B Assuming consider an elementary length \(\mathrm{dx}\) Mass of small element of \(\mathrm{dx}\) length is, \(\mathrm{dm}=\mathrm{kx} . \mathrm{dx}\) [where \(\mathrm{k}\) is the proportionality constant] At the centre of gravity of the \(\operatorname{rod}\left(x_{d}\right)\) is given. \(\mathrm{x}_{\mathrm{d}}=\frac{\int_{0}^{3} \mathrm{x} \cdot \mathrm{dm}}{\int_{0}^{3} \mathrm{dm}}\) \(\left(x_{d}\right)=\frac{\int_{0}^{3} k x \cdot d x \cdot x}{\int_{0}^{3} k x \cdot d x}=\frac{\int_{0}^{3} x^{2} \cdot d x}{\int_{0}^{3} x \cdot d x}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{3}}{\left[\frac{x^{2}}{2}\right]_{0}^{3}}\) Or \(\quad\left(\mathrm{x}_{\mathrm{d}}\right)=\frac{27 / 3}{9 / 2}=2 \mathrm{~m}\)
AIPMT-2002
Rotational Motion
149759
Two bodies of masses \(1 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) have position vectors \(\hat{i}+2 \hat{j}+\hat{k}\) and \(-3 \hat{i}-2 \hat{j}+\hat{k}\), respectively. The centre of mass of this system has a position vector
1 \(-2 \hat{i}+2 \hat{k}\)
2 \(-2 \hat{i}-\hat{j}+\hat{k}\)
3 \(2 \hat{i}-\hat{j}-2 \hat{k}\)
4 \(-\hat{i}+\hat{j}+\hat{k}\)
Explanation:
B Given that, Masses of bodies - \(\mathrm{M}_{1}=1 \mathrm{~kg} \& \mathrm{M}_{2}=3 \mathrm{~kg}\) Position vectors are - \(\overrightarrow{\mathrm{r}}_{1}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{r}}_{2}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) Position vector of the centre of mass \(\vec{r}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{2} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(= \frac{1 \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+3(-3 \hat{i}-2 \hat{j}+\hat{\mathrm{k}})}{1+3}\) \(= \frac{-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{4}=\frac{4(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})}{4}\) \(\overrightarrow{\mathrm{r}}_{\mathrm{cm}}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
AIPMT-2009
Rotational Motion
149761
A rod of weight \(w\) is supported by two parallel knife edges \(A\) and \(B\) and is in equilibrium in a horizontal position. The knives are at a distanced \(d\) from each other. The centre of mass of the rod is at distance \(x\) from \(A\). The normal reaction on \(A\) is
D According to question, situation is shown in below figure- Where, \(N_{A}=\text { Normal reaction on } A\) \(N_{B}=\text { Normal reaction on } B\) \(W=\text { weight of rod }\) In equilibrium position, \(\mathrm{N}_{\mathrm{A}}+\mathrm{N}_{\mathrm{B}}=\mathrm{w}\) By Torque balancing about centre of mass of the rod, \(\mathrm{N}_{\mathrm{A}} \mathrm{x}=\mathrm{N}_{\mathrm{B}}(\mathrm{d}-\mathrm{x})\) Putting the value of \(N_{B}\) from equation (i) \(N_{A} x=\left(w-N_{A}\right)(d-x)\) \(N_{A} x=w d-w x-N_{A} d+N_{A} x\) \(N_{A}=\frac{w(d-x)}{d}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Rotational Motion
149753
A wheel of bicycle is rolling without slipping on a level road. The velocity of the centre of mass is \(\mathbf{v}_{\mathrm{CM}}\), then true statement is
1 The velocity of point \(A\) is \(2 v_{\mathrm{CM}}\) and velocity of point \(B\) is zero
2 The velocity of point \(A\) is zero and velocity of point \(B\) is \(2 \mathrm{v}_{\mathrm{CM}}\)
3 The velocity of point \(A\) is \(2 \mathrm{v}_{\mathrm{CM}}\) and velocity of point \(B\) is \(-v_{C M}\)
4 The velocities of both \(A\) and \(B\) are \(v_{C M}\)
Explanation:
A Let the circle radius is \(\mathrm{R}\) and with \(\omega\). The velocity of centre of mass of body rolling without slipping is given by. \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\) velocity at point \(\mathrm{A}-\) \(\mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{cm}}+\mathrm{R} \omega\) \(=\mathrm{v}_{\mathrm{cm}}+\mathrm{v}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{A}}=2 \mathrm{v}_{\mathrm{cm}}\) velocity at point \(\mathrm{B}-\) \(\mathrm{v}_{\mathrm{B}}=\mathrm{v}_{\mathrm{cm}}-\mathrm{R} \omega\) \(=\mathrm{V}_{\mathrm{cm}}-\mathrm{V}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{B}}=0\)
AIPMT-2001
Rotational Motion
149754
A light rod of length \(l\) has two masses \(m_{1}\) and \(m_{2}\) attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
A To the given data, Length of \(\operatorname{rod}=l\) masses \(=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) From the above figure \(\mathrm{r}_{1} =\frac{\mathrm{m}_{1} \times 0+\mathrm{m}_{2} \times l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{r}_{2} =\frac{\mathrm{m}_{1} \times l+\mathrm{m}_{2} \times 0}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{1} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) Now at the moment of inertia about centre of mass \(\mathrm{I} =\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2}^{2} \mathrm{r}_{2}^{2}\) \(=\mathrm{m}_{1} \times \frac{\mathrm{m}_{2}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}+\mathrm{m}_{2} \times \frac{\mathrm{m}_{1}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\) \(=\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{I} =\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
NEET- 2016
Rotational Motion
149757
A rod is of length \(3 \mathrm{~m}\) and its mass acting per unit length is directly proportional to distance \(x\) from its one end. The centre of gravity of the rod from that end will be at
1 \(1.5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(2.5 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
B Assuming consider an elementary length \(\mathrm{dx}\) Mass of small element of \(\mathrm{dx}\) length is, \(\mathrm{dm}=\mathrm{kx} . \mathrm{dx}\) [where \(\mathrm{k}\) is the proportionality constant] At the centre of gravity of the \(\operatorname{rod}\left(x_{d}\right)\) is given. \(\mathrm{x}_{\mathrm{d}}=\frac{\int_{0}^{3} \mathrm{x} \cdot \mathrm{dm}}{\int_{0}^{3} \mathrm{dm}}\) \(\left(x_{d}\right)=\frac{\int_{0}^{3} k x \cdot d x \cdot x}{\int_{0}^{3} k x \cdot d x}=\frac{\int_{0}^{3} x^{2} \cdot d x}{\int_{0}^{3} x \cdot d x}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{3}}{\left[\frac{x^{2}}{2}\right]_{0}^{3}}\) Or \(\quad\left(\mathrm{x}_{\mathrm{d}}\right)=\frac{27 / 3}{9 / 2}=2 \mathrm{~m}\)
AIPMT-2002
Rotational Motion
149759
Two bodies of masses \(1 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) have position vectors \(\hat{i}+2 \hat{j}+\hat{k}\) and \(-3 \hat{i}-2 \hat{j}+\hat{k}\), respectively. The centre of mass of this system has a position vector
1 \(-2 \hat{i}+2 \hat{k}\)
2 \(-2 \hat{i}-\hat{j}+\hat{k}\)
3 \(2 \hat{i}-\hat{j}-2 \hat{k}\)
4 \(-\hat{i}+\hat{j}+\hat{k}\)
Explanation:
B Given that, Masses of bodies - \(\mathrm{M}_{1}=1 \mathrm{~kg} \& \mathrm{M}_{2}=3 \mathrm{~kg}\) Position vectors are - \(\overrightarrow{\mathrm{r}}_{1}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{r}}_{2}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) Position vector of the centre of mass \(\vec{r}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{2} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(= \frac{1 \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+3(-3 \hat{i}-2 \hat{j}+\hat{\mathrm{k}})}{1+3}\) \(= \frac{-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{4}=\frac{4(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})}{4}\) \(\overrightarrow{\mathrm{r}}_{\mathrm{cm}}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
AIPMT-2009
Rotational Motion
149761
A rod of weight \(w\) is supported by two parallel knife edges \(A\) and \(B\) and is in equilibrium in a horizontal position. The knives are at a distanced \(d\) from each other. The centre of mass of the rod is at distance \(x\) from \(A\). The normal reaction on \(A\) is
D According to question, situation is shown in below figure- Where, \(N_{A}=\text { Normal reaction on } A\) \(N_{B}=\text { Normal reaction on } B\) \(W=\text { weight of rod }\) In equilibrium position, \(\mathrm{N}_{\mathrm{A}}+\mathrm{N}_{\mathrm{B}}=\mathrm{w}\) By Torque balancing about centre of mass of the rod, \(\mathrm{N}_{\mathrm{A}} \mathrm{x}=\mathrm{N}_{\mathrm{B}}(\mathrm{d}-\mathrm{x})\) Putting the value of \(N_{B}\) from equation (i) \(N_{A} x=\left(w-N_{A}\right)(d-x)\) \(N_{A} x=w d-w x-N_{A} d+N_{A} x\) \(N_{A}=\frac{w(d-x)}{d}\)
149753
A wheel of bicycle is rolling without slipping on a level road. The velocity of the centre of mass is \(\mathbf{v}_{\mathrm{CM}}\), then true statement is
1 The velocity of point \(A\) is \(2 v_{\mathrm{CM}}\) and velocity of point \(B\) is zero
2 The velocity of point \(A\) is zero and velocity of point \(B\) is \(2 \mathrm{v}_{\mathrm{CM}}\)
3 The velocity of point \(A\) is \(2 \mathrm{v}_{\mathrm{CM}}\) and velocity of point \(B\) is \(-v_{C M}\)
4 The velocities of both \(A\) and \(B\) are \(v_{C M}\)
Explanation:
A Let the circle radius is \(\mathrm{R}\) and with \(\omega\). The velocity of centre of mass of body rolling without slipping is given by. \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\) velocity at point \(\mathrm{A}-\) \(\mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{cm}}+\mathrm{R} \omega\) \(=\mathrm{v}_{\mathrm{cm}}+\mathrm{v}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{A}}=2 \mathrm{v}_{\mathrm{cm}}\) velocity at point \(\mathrm{B}-\) \(\mathrm{v}_{\mathrm{B}}=\mathrm{v}_{\mathrm{cm}}-\mathrm{R} \omega\) \(=\mathrm{V}_{\mathrm{cm}}-\mathrm{V}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{B}}=0\)
AIPMT-2001
Rotational Motion
149754
A light rod of length \(l\) has two masses \(m_{1}\) and \(m_{2}\) attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
A To the given data, Length of \(\operatorname{rod}=l\) masses \(=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) From the above figure \(\mathrm{r}_{1} =\frac{\mathrm{m}_{1} \times 0+\mathrm{m}_{2} \times l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{r}_{2} =\frac{\mathrm{m}_{1} \times l+\mathrm{m}_{2} \times 0}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{1} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) Now at the moment of inertia about centre of mass \(\mathrm{I} =\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2}^{2} \mathrm{r}_{2}^{2}\) \(=\mathrm{m}_{1} \times \frac{\mathrm{m}_{2}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}+\mathrm{m}_{2} \times \frac{\mathrm{m}_{1}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\) \(=\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{I} =\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
NEET- 2016
Rotational Motion
149757
A rod is of length \(3 \mathrm{~m}\) and its mass acting per unit length is directly proportional to distance \(x\) from its one end. The centre of gravity of the rod from that end will be at
1 \(1.5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(2.5 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
B Assuming consider an elementary length \(\mathrm{dx}\) Mass of small element of \(\mathrm{dx}\) length is, \(\mathrm{dm}=\mathrm{kx} . \mathrm{dx}\) [where \(\mathrm{k}\) is the proportionality constant] At the centre of gravity of the \(\operatorname{rod}\left(x_{d}\right)\) is given. \(\mathrm{x}_{\mathrm{d}}=\frac{\int_{0}^{3} \mathrm{x} \cdot \mathrm{dm}}{\int_{0}^{3} \mathrm{dm}}\) \(\left(x_{d}\right)=\frac{\int_{0}^{3} k x \cdot d x \cdot x}{\int_{0}^{3} k x \cdot d x}=\frac{\int_{0}^{3} x^{2} \cdot d x}{\int_{0}^{3} x \cdot d x}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{3}}{\left[\frac{x^{2}}{2}\right]_{0}^{3}}\) Or \(\quad\left(\mathrm{x}_{\mathrm{d}}\right)=\frac{27 / 3}{9 / 2}=2 \mathrm{~m}\)
AIPMT-2002
Rotational Motion
149759
Two bodies of masses \(1 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) have position vectors \(\hat{i}+2 \hat{j}+\hat{k}\) and \(-3 \hat{i}-2 \hat{j}+\hat{k}\), respectively. The centre of mass of this system has a position vector
1 \(-2 \hat{i}+2 \hat{k}\)
2 \(-2 \hat{i}-\hat{j}+\hat{k}\)
3 \(2 \hat{i}-\hat{j}-2 \hat{k}\)
4 \(-\hat{i}+\hat{j}+\hat{k}\)
Explanation:
B Given that, Masses of bodies - \(\mathrm{M}_{1}=1 \mathrm{~kg} \& \mathrm{M}_{2}=3 \mathrm{~kg}\) Position vectors are - \(\overrightarrow{\mathrm{r}}_{1}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{r}}_{2}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) Position vector of the centre of mass \(\vec{r}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{2} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(= \frac{1 \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+3(-3 \hat{i}-2 \hat{j}+\hat{\mathrm{k}})}{1+3}\) \(= \frac{-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{4}=\frac{4(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})}{4}\) \(\overrightarrow{\mathrm{r}}_{\mathrm{cm}}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
AIPMT-2009
Rotational Motion
149761
A rod of weight \(w\) is supported by two parallel knife edges \(A\) and \(B\) and is in equilibrium in a horizontal position. The knives are at a distanced \(d\) from each other. The centre of mass of the rod is at distance \(x\) from \(A\). The normal reaction on \(A\) is
D According to question, situation is shown in below figure- Where, \(N_{A}=\text { Normal reaction on } A\) \(N_{B}=\text { Normal reaction on } B\) \(W=\text { weight of rod }\) In equilibrium position, \(\mathrm{N}_{\mathrm{A}}+\mathrm{N}_{\mathrm{B}}=\mathrm{w}\) By Torque balancing about centre of mass of the rod, \(\mathrm{N}_{\mathrm{A}} \mathrm{x}=\mathrm{N}_{\mathrm{B}}(\mathrm{d}-\mathrm{x})\) Putting the value of \(N_{B}\) from equation (i) \(N_{A} x=\left(w-N_{A}\right)(d-x)\) \(N_{A} x=w d-w x-N_{A} d+N_{A} x\) \(N_{A}=\frac{w(d-x)}{d}\)
149753
A wheel of bicycle is rolling without slipping on a level road. The velocity of the centre of mass is \(\mathbf{v}_{\mathrm{CM}}\), then true statement is
1 The velocity of point \(A\) is \(2 v_{\mathrm{CM}}\) and velocity of point \(B\) is zero
2 The velocity of point \(A\) is zero and velocity of point \(B\) is \(2 \mathrm{v}_{\mathrm{CM}}\)
3 The velocity of point \(A\) is \(2 \mathrm{v}_{\mathrm{CM}}\) and velocity of point \(B\) is \(-v_{C M}\)
4 The velocities of both \(A\) and \(B\) are \(v_{C M}\)
Explanation:
A Let the circle radius is \(\mathrm{R}\) and with \(\omega\). The velocity of centre of mass of body rolling without slipping is given by. \(\mathrm{v}_{\mathrm{cm}}=\mathrm{R} \omega\) velocity at point \(\mathrm{A}-\) \(\mathrm{v}_{\mathrm{A}}=\mathrm{v}_{\mathrm{cm}}+\mathrm{R} \omega\) \(=\mathrm{v}_{\mathrm{cm}}+\mathrm{v}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{A}}=2 \mathrm{v}_{\mathrm{cm}}\) velocity at point \(\mathrm{B}-\) \(\mathrm{v}_{\mathrm{B}}=\mathrm{v}_{\mathrm{cm}}-\mathrm{R} \omega\) \(=\mathrm{V}_{\mathrm{cm}}-\mathrm{V}_{\mathrm{cm}}\) \(\mathrm{v}_{\mathrm{B}}=0\)
AIPMT-2001
Rotational Motion
149754
A light rod of length \(l\) has two masses \(m_{1}\) and \(m_{2}\) attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is
A To the given data, Length of \(\operatorname{rod}=l\) masses \(=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\) From the above figure \(\mathrm{r}_{1} =\frac{\mathrm{m}_{1} \times 0+\mathrm{m}_{2} \times l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{2} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(\mathrm{r}_{2} =\frac{\mathrm{m}_{1} \times l+\mathrm{m}_{2} \times 0}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(=\frac{\mathrm{m}_{1} l}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) Now at the moment of inertia about centre of mass \(\mathrm{I} =\mathrm{m}_{1} \mathrm{r}_{1}^{2}+\mathrm{m}_{2}^{2} \mathrm{r}_{2}^{2}\) \(=\mathrm{m}_{1} \times \frac{\mathrm{m}_{2}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}+\mathrm{m}_{2} \times \frac{\mathrm{m}_{1}{ }^{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\) \(=\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)^{2}}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)\) \(\mathrm{I} =\frac{\mathrm{m}_{1} \mathrm{~m}_{2} l^{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\)
NEET- 2016
Rotational Motion
149757
A rod is of length \(3 \mathrm{~m}\) and its mass acting per unit length is directly proportional to distance \(x\) from its one end. The centre of gravity of the rod from that end will be at
1 \(1.5 \mathrm{~m}\)
2 \(2 \mathrm{~m}\)
3 \(2.5 \mathrm{~m}\)
4 \(3 \mathrm{~m}\)
Explanation:
B Assuming consider an elementary length \(\mathrm{dx}\) Mass of small element of \(\mathrm{dx}\) length is, \(\mathrm{dm}=\mathrm{kx} . \mathrm{dx}\) [where \(\mathrm{k}\) is the proportionality constant] At the centre of gravity of the \(\operatorname{rod}\left(x_{d}\right)\) is given. \(\mathrm{x}_{\mathrm{d}}=\frac{\int_{0}^{3} \mathrm{x} \cdot \mathrm{dm}}{\int_{0}^{3} \mathrm{dm}}\) \(\left(x_{d}\right)=\frac{\int_{0}^{3} k x \cdot d x \cdot x}{\int_{0}^{3} k x \cdot d x}=\frac{\int_{0}^{3} x^{2} \cdot d x}{\int_{0}^{3} x \cdot d x}=\frac{\left[\frac{x^{3}}{3}\right]_{0}^{3}}{\left[\frac{x^{2}}{2}\right]_{0}^{3}}\) Or \(\quad\left(\mathrm{x}_{\mathrm{d}}\right)=\frac{27 / 3}{9 / 2}=2 \mathrm{~m}\)
AIPMT-2002
Rotational Motion
149759
Two bodies of masses \(1 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) have position vectors \(\hat{i}+2 \hat{j}+\hat{k}\) and \(-3 \hat{i}-2 \hat{j}+\hat{k}\), respectively. The centre of mass of this system has a position vector
1 \(-2 \hat{i}+2 \hat{k}\)
2 \(-2 \hat{i}-\hat{j}+\hat{k}\)
3 \(2 \hat{i}-\hat{j}-2 \hat{k}\)
4 \(-\hat{i}+\hat{j}+\hat{k}\)
Explanation:
B Given that, Masses of bodies - \(\mathrm{M}_{1}=1 \mathrm{~kg} \& \mathrm{M}_{2}=3 \mathrm{~kg}\) Position vectors are - \(\overrightarrow{\mathrm{r}}_{1}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{r}}_{2}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) Position vector of the centre of mass \(\vec{r}_{\mathrm{cm}}=\frac{\mathrm{m}_{1} \overrightarrow{\mathrm{r}}_{1}+\mathrm{m}_{2} \overrightarrow{\mathrm{r}}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\) \(= \frac{1 \times(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+3(-3 \hat{i}-2 \hat{j}+\hat{\mathrm{k}})}{1+3}\) \(= \frac{-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}}{4}=\frac{4(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})}{4}\) \(\overrightarrow{\mathrm{r}}_{\mathrm{cm}}=-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
AIPMT-2009
Rotational Motion
149761
A rod of weight \(w\) is supported by two parallel knife edges \(A\) and \(B\) and is in equilibrium in a horizontal position. The knives are at a distanced \(d\) from each other. The centre of mass of the rod is at distance \(x\) from \(A\). The normal reaction on \(A\) is
D According to question, situation is shown in below figure- Where, \(N_{A}=\text { Normal reaction on } A\) \(N_{B}=\text { Normal reaction on } B\) \(W=\text { weight of rod }\) In equilibrium position, \(\mathrm{N}_{\mathrm{A}}+\mathrm{N}_{\mathrm{B}}=\mathrm{w}\) By Torque balancing about centre of mass of the rod, \(\mathrm{N}_{\mathrm{A}} \mathrm{x}=\mathrm{N}_{\mathrm{B}}(\mathrm{d}-\mathrm{x})\) Putting the value of \(N_{B}\) from equation (i) \(N_{A} x=\left(w-N_{A}\right)(d-x)\) \(N_{A} x=w d-w x-N_{A} d+N_{A} x\) \(N_{A}=\frac{w(d-x)}{d}\)
