149821
A particle moves along a circle of radius \(\frac{20}{\pi} \mathrm{m}\) with constant tangential acceleration. If the velocity of the particle is \(80 \mathrm{~m} \mathrm{~s}^{-1}\) at the end of the second revolution after motion has begun, the tangential acceleration is
1 \(640 \pi \mathrm{m} \mathrm{s}^{-2}\)
2 \(140 \pi \mathrm{m} \mathrm{s}^{-2}\)
3 \(40 \pi \mathrm{m} \mathrm{s}^{-2}\)
4 \(40 \mathrm{~m} \mathrm{~s}^{-2}\)
Explanation:
D Given that, \(\mathrm{r}=\frac{20}{\pi}, \mathrm{v}=80 \mathrm{~m} / \mathrm{sec}\) In two revolution, Distance \((\mathrm{s})=2 \times 2 \pi \mathrm{r}=2 \times 2 \times \pi \times \frac{20}{\pi}=80 \mathrm{~m}\) We know \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \quad \text { (initial velocity is zero) }\) \(\mathrm{v}^{2}=2 \mathrm{as}\) \(\mathrm{a}=\frac{\mathrm{v}^{2}}{2 \mathrm{~s}}\) \(\mathrm{a}=\frac{80 \times 80}{2 \times 80}\) \(\mathrm{a}=40 \mathrm{~m} / \mathrm{sec}^{2}\) \(\because \quad \mathrm{v}^{2}=2 \mathrm{as}\)
COMEDK 2019
Rotational Motion
149822
A uniform rod of length \(l\) and density \(\rho\) is revolving about a vertical axis passing through its one end. If \(\omega\) is the angular velocity of the rod then the centrifugal force per unit area of the rod is
1 \(\frac{\rho \omega^{2} l^{2}}{4}\)
2 \(\frac{\rho \omega^{2} l^{2}}{12}\)
3 \(\frac{\rho \omega^{2} l^{2}}{2}\)
4 \(\frac{\rho \omega^{2} l^{2}}{8}\)
Explanation:
C Centrifugal force per unit area of rod for small length \(d x\) is \(\int_{0}^{\mathrm{F}} \mathrm{dF}=\int_{0}^{l} \frac{\mathrm{m} \omega^{2} \mathrm{dx}}{\mathrm{A}}\) Let \(A\) is the area of rod and \(x\) is variable length of the rod. Then density of rod, \(\rho=\frac{\mathrm{m}}{\mathrm{A} \cdot \mathrm{x}} \text { or } \mathrm{m}=\rho \mathrm{Ax}\) putting the value of \(m\) in equation (i), we get \(\mathrm{F}=\rho \frac{\omega^{2}}{\mathrm{~A}} \int_{0}^{l} \mathrm{~A} . \mathrm{x} . \mathrm{dx}\) \(\mathrm{F}=\rho \omega^{2}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{l}=\frac{\rho \omega^{2} l^{2}}{2}\) \(\mathrm{~F}=\frac{\rho \omega^{2} l^{2}}{2}\)
AP EAMCET (21.04.2019) Shift-II
Rotational Motion
149824
An elevator is going up with an acceleration 2 \(\mathrm{m} / \mathrm{s}^{2}\). If radius of the wheel attached to the elevator is \(0.1 \mathrm{~m}\), then find out number of revolutions in \(\mathbf{t}=10 \mathrm{~s}\).
1 129
2 139
3 159
4 179
Explanation:
C Given that, \(\mathrm{a}=2 \mathrm{~m} / \mathrm{sec}^{2}\) \(\mathrm{r}=0.1 \mathrm{~m}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\theta=2 \pi \mathrm{n}\) and \(\quad \theta=\frac{1}{2} \alpha \mathrm{t}^{2}\) [If initial speed is zero] \(\mathrm{n}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{r}}\right) \frac{\mathrm{t}^{2}}{2 \pi} \quad\left[\because \alpha=\frac{\mathrm{a}}{\mathrm{r}}\right]\) \(\mathrm{n}=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{(10)^{2}}{6.28}\) \(\mathrm{n}=159.23\) Number of revolution, \(\mathrm{n}=159\)
AIIMS-27.05.2018(M)
Rotational Motion
149825
Consider a wheel rotating around a fixed axis. If the rotation angle \(\theta\) varies with time as \(\theta=\) \(\mathrm{at}^{2}\), then the total acceleration of a point \(A\) on the rim of the wheel is ( \(\mathrm{v}\) being the tangential velocity)
C Given, \(\theta=\mathrm{at}^{2}\) Angular velocity \((\omega)=\frac{\mathrm{d} \theta}{\mathrm{dt}}=2 \mathrm{at}\) Angular acceleration \((\alpha)=\frac{\mathrm{d} \omega}{\mathrm{dt}}=2 \mathrm{a}\) Tangential linear acceleration, \(a_{\mathrm{t}} =\alpha \mathrm{R}\) \(\mathrm{a} =2 \mathrm{aR}\) Radial or normal acceleration of particle \(a_{r} =\omega^{2} R\) \(=(2 a t)^{2} R\) \(a_{r} =4 a^{2} t^{2} R\) From equation (i) and (ii) \(\mathrm{a}_{\text {total }} =\sqrt{\mathrm{a}_{\mathrm{t}}^{2}+\mathrm{a}_{\mathrm{r}}^{2}}=\sqrt{(2 \mathrm{aR})^{2}+\left(4 \mathrm{a}^{2} \mathrm{t}^{2} \mathrm{R}\right)^{2}}\) \(\mathrm{a}_{\text {total }} =2 \mathrm{aR} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}} \quad\left[\because 2 \mathrm{aR}=\frac{\mathrm{v}}{\mathrm{t}}\right]\) \(\therefore \quad \mathrm{a}_{\text {total }} =\frac{\mathrm{v}}{\mathrm{t}} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}}\)
TS- EAMCET-05.05.2018
Rotational Motion
149826
The angular speed of second's hand of the clock is
1 \(\frac{\pi}{1800} \operatorname{rad~s}^{-1}\)
2 \(\frac{\pi}{30} \operatorname{rad~s}^{-1}\)
3 \(\frac{\pi}{90} \mathrm{rads}^{-1}\)
4 \(\frac{\pi}{60} \mathrm{rads}^{-1}\)
Explanation:
B Given, Speed of second hand of the clock \(\mathrm{t}=60 \mathrm{sec}\). \(\theta=2 \pi \mathrm{rad}\) Angular speed \((\omega)=\frac{\text { Angular displacement }(\theta)}{\operatorname{time}(\mathrm{t})}\) \(=\frac{2 \pi}{60}\) \(\omega=\frac{\pi}{30} \mathrm{rad} / \mathrm{sec}\)
149821
A particle moves along a circle of radius \(\frac{20}{\pi} \mathrm{m}\) with constant tangential acceleration. If the velocity of the particle is \(80 \mathrm{~m} \mathrm{~s}^{-1}\) at the end of the second revolution after motion has begun, the tangential acceleration is
1 \(640 \pi \mathrm{m} \mathrm{s}^{-2}\)
2 \(140 \pi \mathrm{m} \mathrm{s}^{-2}\)
3 \(40 \pi \mathrm{m} \mathrm{s}^{-2}\)
4 \(40 \mathrm{~m} \mathrm{~s}^{-2}\)
Explanation:
D Given that, \(\mathrm{r}=\frac{20}{\pi}, \mathrm{v}=80 \mathrm{~m} / \mathrm{sec}\) In two revolution, Distance \((\mathrm{s})=2 \times 2 \pi \mathrm{r}=2 \times 2 \times \pi \times \frac{20}{\pi}=80 \mathrm{~m}\) We know \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \quad \text { (initial velocity is zero) }\) \(\mathrm{v}^{2}=2 \mathrm{as}\) \(\mathrm{a}=\frac{\mathrm{v}^{2}}{2 \mathrm{~s}}\) \(\mathrm{a}=\frac{80 \times 80}{2 \times 80}\) \(\mathrm{a}=40 \mathrm{~m} / \mathrm{sec}^{2}\) \(\because \quad \mathrm{v}^{2}=2 \mathrm{as}\)
COMEDK 2019
Rotational Motion
149822
A uniform rod of length \(l\) and density \(\rho\) is revolving about a vertical axis passing through its one end. If \(\omega\) is the angular velocity of the rod then the centrifugal force per unit area of the rod is
1 \(\frac{\rho \omega^{2} l^{2}}{4}\)
2 \(\frac{\rho \omega^{2} l^{2}}{12}\)
3 \(\frac{\rho \omega^{2} l^{2}}{2}\)
4 \(\frac{\rho \omega^{2} l^{2}}{8}\)
Explanation:
C Centrifugal force per unit area of rod for small length \(d x\) is \(\int_{0}^{\mathrm{F}} \mathrm{dF}=\int_{0}^{l} \frac{\mathrm{m} \omega^{2} \mathrm{dx}}{\mathrm{A}}\) Let \(A\) is the area of rod and \(x\) is variable length of the rod. Then density of rod, \(\rho=\frac{\mathrm{m}}{\mathrm{A} \cdot \mathrm{x}} \text { or } \mathrm{m}=\rho \mathrm{Ax}\) putting the value of \(m\) in equation (i), we get \(\mathrm{F}=\rho \frac{\omega^{2}}{\mathrm{~A}} \int_{0}^{l} \mathrm{~A} . \mathrm{x} . \mathrm{dx}\) \(\mathrm{F}=\rho \omega^{2}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{l}=\frac{\rho \omega^{2} l^{2}}{2}\) \(\mathrm{~F}=\frac{\rho \omega^{2} l^{2}}{2}\)
AP EAMCET (21.04.2019) Shift-II
Rotational Motion
149824
An elevator is going up with an acceleration 2 \(\mathrm{m} / \mathrm{s}^{2}\). If radius of the wheel attached to the elevator is \(0.1 \mathrm{~m}\), then find out number of revolutions in \(\mathbf{t}=10 \mathrm{~s}\).
1 129
2 139
3 159
4 179
Explanation:
C Given that, \(\mathrm{a}=2 \mathrm{~m} / \mathrm{sec}^{2}\) \(\mathrm{r}=0.1 \mathrm{~m}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\theta=2 \pi \mathrm{n}\) and \(\quad \theta=\frac{1}{2} \alpha \mathrm{t}^{2}\) [If initial speed is zero] \(\mathrm{n}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{r}}\right) \frac{\mathrm{t}^{2}}{2 \pi} \quad\left[\because \alpha=\frac{\mathrm{a}}{\mathrm{r}}\right]\) \(\mathrm{n}=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{(10)^{2}}{6.28}\) \(\mathrm{n}=159.23\) Number of revolution, \(\mathrm{n}=159\)
AIIMS-27.05.2018(M)
Rotational Motion
149825
Consider a wheel rotating around a fixed axis. If the rotation angle \(\theta\) varies with time as \(\theta=\) \(\mathrm{at}^{2}\), then the total acceleration of a point \(A\) on the rim of the wheel is ( \(\mathrm{v}\) being the tangential velocity)
C Given, \(\theta=\mathrm{at}^{2}\) Angular velocity \((\omega)=\frac{\mathrm{d} \theta}{\mathrm{dt}}=2 \mathrm{at}\) Angular acceleration \((\alpha)=\frac{\mathrm{d} \omega}{\mathrm{dt}}=2 \mathrm{a}\) Tangential linear acceleration, \(a_{\mathrm{t}} =\alpha \mathrm{R}\) \(\mathrm{a} =2 \mathrm{aR}\) Radial or normal acceleration of particle \(a_{r} =\omega^{2} R\) \(=(2 a t)^{2} R\) \(a_{r} =4 a^{2} t^{2} R\) From equation (i) and (ii) \(\mathrm{a}_{\text {total }} =\sqrt{\mathrm{a}_{\mathrm{t}}^{2}+\mathrm{a}_{\mathrm{r}}^{2}}=\sqrt{(2 \mathrm{aR})^{2}+\left(4 \mathrm{a}^{2} \mathrm{t}^{2} \mathrm{R}\right)^{2}}\) \(\mathrm{a}_{\text {total }} =2 \mathrm{aR} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}} \quad\left[\because 2 \mathrm{aR}=\frac{\mathrm{v}}{\mathrm{t}}\right]\) \(\therefore \quad \mathrm{a}_{\text {total }} =\frac{\mathrm{v}}{\mathrm{t}} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}}\)
TS- EAMCET-05.05.2018
Rotational Motion
149826
The angular speed of second's hand of the clock is
1 \(\frac{\pi}{1800} \operatorname{rad~s}^{-1}\)
2 \(\frac{\pi}{30} \operatorname{rad~s}^{-1}\)
3 \(\frac{\pi}{90} \mathrm{rads}^{-1}\)
4 \(\frac{\pi}{60} \mathrm{rads}^{-1}\)
Explanation:
B Given, Speed of second hand of the clock \(\mathrm{t}=60 \mathrm{sec}\). \(\theta=2 \pi \mathrm{rad}\) Angular speed \((\omega)=\frac{\text { Angular displacement }(\theta)}{\operatorname{time}(\mathrm{t})}\) \(=\frac{2 \pi}{60}\) \(\omega=\frac{\pi}{30} \mathrm{rad} / \mathrm{sec}\)
149821
A particle moves along a circle of radius \(\frac{20}{\pi} \mathrm{m}\) with constant tangential acceleration. If the velocity of the particle is \(80 \mathrm{~m} \mathrm{~s}^{-1}\) at the end of the second revolution after motion has begun, the tangential acceleration is
1 \(640 \pi \mathrm{m} \mathrm{s}^{-2}\)
2 \(140 \pi \mathrm{m} \mathrm{s}^{-2}\)
3 \(40 \pi \mathrm{m} \mathrm{s}^{-2}\)
4 \(40 \mathrm{~m} \mathrm{~s}^{-2}\)
Explanation:
D Given that, \(\mathrm{r}=\frac{20}{\pi}, \mathrm{v}=80 \mathrm{~m} / \mathrm{sec}\) In two revolution, Distance \((\mathrm{s})=2 \times 2 \pi \mathrm{r}=2 \times 2 \times \pi \times \frac{20}{\pi}=80 \mathrm{~m}\) We know \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \quad \text { (initial velocity is zero) }\) \(\mathrm{v}^{2}=2 \mathrm{as}\) \(\mathrm{a}=\frac{\mathrm{v}^{2}}{2 \mathrm{~s}}\) \(\mathrm{a}=\frac{80 \times 80}{2 \times 80}\) \(\mathrm{a}=40 \mathrm{~m} / \mathrm{sec}^{2}\) \(\because \quad \mathrm{v}^{2}=2 \mathrm{as}\)
COMEDK 2019
Rotational Motion
149822
A uniform rod of length \(l\) and density \(\rho\) is revolving about a vertical axis passing through its one end. If \(\omega\) is the angular velocity of the rod then the centrifugal force per unit area of the rod is
1 \(\frac{\rho \omega^{2} l^{2}}{4}\)
2 \(\frac{\rho \omega^{2} l^{2}}{12}\)
3 \(\frac{\rho \omega^{2} l^{2}}{2}\)
4 \(\frac{\rho \omega^{2} l^{2}}{8}\)
Explanation:
C Centrifugal force per unit area of rod for small length \(d x\) is \(\int_{0}^{\mathrm{F}} \mathrm{dF}=\int_{0}^{l} \frac{\mathrm{m} \omega^{2} \mathrm{dx}}{\mathrm{A}}\) Let \(A\) is the area of rod and \(x\) is variable length of the rod. Then density of rod, \(\rho=\frac{\mathrm{m}}{\mathrm{A} \cdot \mathrm{x}} \text { or } \mathrm{m}=\rho \mathrm{Ax}\) putting the value of \(m\) in equation (i), we get \(\mathrm{F}=\rho \frac{\omega^{2}}{\mathrm{~A}} \int_{0}^{l} \mathrm{~A} . \mathrm{x} . \mathrm{dx}\) \(\mathrm{F}=\rho \omega^{2}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{l}=\frac{\rho \omega^{2} l^{2}}{2}\) \(\mathrm{~F}=\frac{\rho \omega^{2} l^{2}}{2}\)
AP EAMCET (21.04.2019) Shift-II
Rotational Motion
149824
An elevator is going up with an acceleration 2 \(\mathrm{m} / \mathrm{s}^{2}\). If radius of the wheel attached to the elevator is \(0.1 \mathrm{~m}\), then find out number of revolutions in \(\mathbf{t}=10 \mathrm{~s}\).
1 129
2 139
3 159
4 179
Explanation:
C Given that, \(\mathrm{a}=2 \mathrm{~m} / \mathrm{sec}^{2}\) \(\mathrm{r}=0.1 \mathrm{~m}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\theta=2 \pi \mathrm{n}\) and \(\quad \theta=\frac{1}{2} \alpha \mathrm{t}^{2}\) [If initial speed is zero] \(\mathrm{n}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{r}}\right) \frac{\mathrm{t}^{2}}{2 \pi} \quad\left[\because \alpha=\frac{\mathrm{a}}{\mathrm{r}}\right]\) \(\mathrm{n}=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{(10)^{2}}{6.28}\) \(\mathrm{n}=159.23\) Number of revolution, \(\mathrm{n}=159\)
AIIMS-27.05.2018(M)
Rotational Motion
149825
Consider a wheel rotating around a fixed axis. If the rotation angle \(\theta\) varies with time as \(\theta=\) \(\mathrm{at}^{2}\), then the total acceleration of a point \(A\) on the rim of the wheel is ( \(\mathrm{v}\) being the tangential velocity)
C Given, \(\theta=\mathrm{at}^{2}\) Angular velocity \((\omega)=\frac{\mathrm{d} \theta}{\mathrm{dt}}=2 \mathrm{at}\) Angular acceleration \((\alpha)=\frac{\mathrm{d} \omega}{\mathrm{dt}}=2 \mathrm{a}\) Tangential linear acceleration, \(a_{\mathrm{t}} =\alpha \mathrm{R}\) \(\mathrm{a} =2 \mathrm{aR}\) Radial or normal acceleration of particle \(a_{r} =\omega^{2} R\) \(=(2 a t)^{2} R\) \(a_{r} =4 a^{2} t^{2} R\) From equation (i) and (ii) \(\mathrm{a}_{\text {total }} =\sqrt{\mathrm{a}_{\mathrm{t}}^{2}+\mathrm{a}_{\mathrm{r}}^{2}}=\sqrt{(2 \mathrm{aR})^{2}+\left(4 \mathrm{a}^{2} \mathrm{t}^{2} \mathrm{R}\right)^{2}}\) \(\mathrm{a}_{\text {total }} =2 \mathrm{aR} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}} \quad\left[\because 2 \mathrm{aR}=\frac{\mathrm{v}}{\mathrm{t}}\right]\) \(\therefore \quad \mathrm{a}_{\text {total }} =\frac{\mathrm{v}}{\mathrm{t}} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}}\)
TS- EAMCET-05.05.2018
Rotational Motion
149826
The angular speed of second's hand of the clock is
1 \(\frac{\pi}{1800} \operatorname{rad~s}^{-1}\)
2 \(\frac{\pi}{30} \operatorname{rad~s}^{-1}\)
3 \(\frac{\pi}{90} \mathrm{rads}^{-1}\)
4 \(\frac{\pi}{60} \mathrm{rads}^{-1}\)
Explanation:
B Given, Speed of second hand of the clock \(\mathrm{t}=60 \mathrm{sec}\). \(\theta=2 \pi \mathrm{rad}\) Angular speed \((\omega)=\frac{\text { Angular displacement }(\theta)}{\operatorname{time}(\mathrm{t})}\) \(=\frac{2 \pi}{60}\) \(\omega=\frac{\pi}{30} \mathrm{rad} / \mathrm{sec}\)
149821
A particle moves along a circle of radius \(\frac{20}{\pi} \mathrm{m}\) with constant tangential acceleration. If the velocity of the particle is \(80 \mathrm{~m} \mathrm{~s}^{-1}\) at the end of the second revolution after motion has begun, the tangential acceleration is
1 \(640 \pi \mathrm{m} \mathrm{s}^{-2}\)
2 \(140 \pi \mathrm{m} \mathrm{s}^{-2}\)
3 \(40 \pi \mathrm{m} \mathrm{s}^{-2}\)
4 \(40 \mathrm{~m} \mathrm{~s}^{-2}\)
Explanation:
D Given that, \(\mathrm{r}=\frac{20}{\pi}, \mathrm{v}=80 \mathrm{~m} / \mathrm{sec}\) In two revolution, Distance \((\mathrm{s})=2 \times 2 \pi \mathrm{r}=2 \times 2 \times \pi \times \frac{20}{\pi}=80 \mathrm{~m}\) We know \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \quad \text { (initial velocity is zero) }\) \(\mathrm{v}^{2}=2 \mathrm{as}\) \(\mathrm{a}=\frac{\mathrm{v}^{2}}{2 \mathrm{~s}}\) \(\mathrm{a}=\frac{80 \times 80}{2 \times 80}\) \(\mathrm{a}=40 \mathrm{~m} / \mathrm{sec}^{2}\) \(\because \quad \mathrm{v}^{2}=2 \mathrm{as}\)
COMEDK 2019
Rotational Motion
149822
A uniform rod of length \(l\) and density \(\rho\) is revolving about a vertical axis passing through its one end. If \(\omega\) is the angular velocity of the rod then the centrifugal force per unit area of the rod is
1 \(\frac{\rho \omega^{2} l^{2}}{4}\)
2 \(\frac{\rho \omega^{2} l^{2}}{12}\)
3 \(\frac{\rho \omega^{2} l^{2}}{2}\)
4 \(\frac{\rho \omega^{2} l^{2}}{8}\)
Explanation:
C Centrifugal force per unit area of rod for small length \(d x\) is \(\int_{0}^{\mathrm{F}} \mathrm{dF}=\int_{0}^{l} \frac{\mathrm{m} \omega^{2} \mathrm{dx}}{\mathrm{A}}\) Let \(A\) is the area of rod and \(x\) is variable length of the rod. Then density of rod, \(\rho=\frac{\mathrm{m}}{\mathrm{A} \cdot \mathrm{x}} \text { or } \mathrm{m}=\rho \mathrm{Ax}\) putting the value of \(m\) in equation (i), we get \(\mathrm{F}=\rho \frac{\omega^{2}}{\mathrm{~A}} \int_{0}^{l} \mathrm{~A} . \mathrm{x} . \mathrm{dx}\) \(\mathrm{F}=\rho \omega^{2}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{l}=\frac{\rho \omega^{2} l^{2}}{2}\) \(\mathrm{~F}=\frac{\rho \omega^{2} l^{2}}{2}\)
AP EAMCET (21.04.2019) Shift-II
Rotational Motion
149824
An elevator is going up with an acceleration 2 \(\mathrm{m} / \mathrm{s}^{2}\). If radius of the wheel attached to the elevator is \(0.1 \mathrm{~m}\), then find out number of revolutions in \(\mathbf{t}=10 \mathrm{~s}\).
1 129
2 139
3 159
4 179
Explanation:
C Given that, \(\mathrm{a}=2 \mathrm{~m} / \mathrm{sec}^{2}\) \(\mathrm{r}=0.1 \mathrm{~m}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\theta=2 \pi \mathrm{n}\) and \(\quad \theta=\frac{1}{2} \alpha \mathrm{t}^{2}\) [If initial speed is zero] \(\mathrm{n}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{r}}\right) \frac{\mathrm{t}^{2}}{2 \pi} \quad\left[\because \alpha=\frac{\mathrm{a}}{\mathrm{r}}\right]\) \(\mathrm{n}=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{(10)^{2}}{6.28}\) \(\mathrm{n}=159.23\) Number of revolution, \(\mathrm{n}=159\)
AIIMS-27.05.2018(M)
Rotational Motion
149825
Consider a wheel rotating around a fixed axis. If the rotation angle \(\theta\) varies with time as \(\theta=\) \(\mathrm{at}^{2}\), then the total acceleration of a point \(A\) on the rim of the wheel is ( \(\mathrm{v}\) being the tangential velocity)
C Given, \(\theta=\mathrm{at}^{2}\) Angular velocity \((\omega)=\frac{\mathrm{d} \theta}{\mathrm{dt}}=2 \mathrm{at}\) Angular acceleration \((\alpha)=\frac{\mathrm{d} \omega}{\mathrm{dt}}=2 \mathrm{a}\) Tangential linear acceleration, \(a_{\mathrm{t}} =\alpha \mathrm{R}\) \(\mathrm{a} =2 \mathrm{aR}\) Radial or normal acceleration of particle \(a_{r} =\omega^{2} R\) \(=(2 a t)^{2} R\) \(a_{r} =4 a^{2} t^{2} R\) From equation (i) and (ii) \(\mathrm{a}_{\text {total }} =\sqrt{\mathrm{a}_{\mathrm{t}}^{2}+\mathrm{a}_{\mathrm{r}}^{2}}=\sqrt{(2 \mathrm{aR})^{2}+\left(4 \mathrm{a}^{2} \mathrm{t}^{2} \mathrm{R}\right)^{2}}\) \(\mathrm{a}_{\text {total }} =2 \mathrm{aR} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}} \quad\left[\because 2 \mathrm{aR}=\frac{\mathrm{v}}{\mathrm{t}}\right]\) \(\therefore \quad \mathrm{a}_{\text {total }} =\frac{\mathrm{v}}{\mathrm{t}} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}}\)
TS- EAMCET-05.05.2018
Rotational Motion
149826
The angular speed of second's hand of the clock is
1 \(\frac{\pi}{1800} \operatorname{rad~s}^{-1}\)
2 \(\frac{\pi}{30} \operatorname{rad~s}^{-1}\)
3 \(\frac{\pi}{90} \mathrm{rads}^{-1}\)
4 \(\frac{\pi}{60} \mathrm{rads}^{-1}\)
Explanation:
B Given, Speed of second hand of the clock \(\mathrm{t}=60 \mathrm{sec}\). \(\theta=2 \pi \mathrm{rad}\) Angular speed \((\omega)=\frac{\text { Angular displacement }(\theta)}{\operatorname{time}(\mathrm{t})}\) \(=\frac{2 \pi}{60}\) \(\omega=\frac{\pi}{30} \mathrm{rad} / \mathrm{sec}\)
149821
A particle moves along a circle of radius \(\frac{20}{\pi} \mathrm{m}\) with constant tangential acceleration. If the velocity of the particle is \(80 \mathrm{~m} \mathrm{~s}^{-1}\) at the end of the second revolution after motion has begun, the tangential acceleration is
1 \(640 \pi \mathrm{m} \mathrm{s}^{-2}\)
2 \(140 \pi \mathrm{m} \mathrm{s}^{-2}\)
3 \(40 \pi \mathrm{m} \mathrm{s}^{-2}\)
4 \(40 \mathrm{~m} \mathrm{~s}^{-2}\)
Explanation:
D Given that, \(\mathrm{r}=\frac{20}{\pi}, \mathrm{v}=80 \mathrm{~m} / \mathrm{sec}\) In two revolution, Distance \((\mathrm{s})=2 \times 2 \pi \mathrm{r}=2 \times 2 \times \pi \times \frac{20}{\pi}=80 \mathrm{~m}\) We know \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{as} \quad \text { (initial velocity is zero) }\) \(\mathrm{v}^{2}=2 \mathrm{as}\) \(\mathrm{a}=\frac{\mathrm{v}^{2}}{2 \mathrm{~s}}\) \(\mathrm{a}=\frac{80 \times 80}{2 \times 80}\) \(\mathrm{a}=40 \mathrm{~m} / \mathrm{sec}^{2}\) \(\because \quad \mathrm{v}^{2}=2 \mathrm{as}\)
COMEDK 2019
Rotational Motion
149822
A uniform rod of length \(l\) and density \(\rho\) is revolving about a vertical axis passing through its one end. If \(\omega\) is the angular velocity of the rod then the centrifugal force per unit area of the rod is
1 \(\frac{\rho \omega^{2} l^{2}}{4}\)
2 \(\frac{\rho \omega^{2} l^{2}}{12}\)
3 \(\frac{\rho \omega^{2} l^{2}}{2}\)
4 \(\frac{\rho \omega^{2} l^{2}}{8}\)
Explanation:
C Centrifugal force per unit area of rod for small length \(d x\) is \(\int_{0}^{\mathrm{F}} \mathrm{dF}=\int_{0}^{l} \frac{\mathrm{m} \omega^{2} \mathrm{dx}}{\mathrm{A}}\) Let \(A\) is the area of rod and \(x\) is variable length of the rod. Then density of rod, \(\rho=\frac{\mathrm{m}}{\mathrm{A} \cdot \mathrm{x}} \text { or } \mathrm{m}=\rho \mathrm{Ax}\) putting the value of \(m\) in equation (i), we get \(\mathrm{F}=\rho \frac{\omega^{2}}{\mathrm{~A}} \int_{0}^{l} \mathrm{~A} . \mathrm{x} . \mathrm{dx}\) \(\mathrm{F}=\rho \omega^{2}\left(\frac{\mathrm{x}^{2}}{2}\right)_{0}^{l}=\frac{\rho \omega^{2} l^{2}}{2}\) \(\mathrm{~F}=\frac{\rho \omega^{2} l^{2}}{2}\)
AP EAMCET (21.04.2019) Shift-II
Rotational Motion
149824
An elevator is going up with an acceleration 2 \(\mathrm{m} / \mathrm{s}^{2}\). If radius of the wheel attached to the elevator is \(0.1 \mathrm{~m}\), then find out number of revolutions in \(\mathbf{t}=10 \mathrm{~s}\).
1 129
2 139
3 159
4 179
Explanation:
C Given that, \(\mathrm{a}=2 \mathrm{~m} / \mathrm{sec}^{2}\) \(\mathrm{r}=0.1 \mathrm{~m}\) \(\mathrm{t}=10 \mathrm{sec}\) We know, \(\theta=2 \pi \mathrm{n}\) and \(\quad \theta=\frac{1}{2} \alpha \mathrm{t}^{2}\) [If initial speed is zero] \(\mathrm{n}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{r}}\right) \frac{\mathrm{t}^{2}}{2 \pi} \quad\left[\because \alpha=\frac{\mathrm{a}}{\mathrm{r}}\right]\) \(\mathrm{n}=\frac{1}{2}\left(\frac{2}{0.1}\right) \frac{(10)^{2}}{6.28}\) \(\mathrm{n}=159.23\) Number of revolution, \(\mathrm{n}=159\)
AIIMS-27.05.2018(M)
Rotational Motion
149825
Consider a wheel rotating around a fixed axis. If the rotation angle \(\theta\) varies with time as \(\theta=\) \(\mathrm{at}^{2}\), then the total acceleration of a point \(A\) on the rim of the wheel is ( \(\mathrm{v}\) being the tangential velocity)
C Given, \(\theta=\mathrm{at}^{2}\) Angular velocity \((\omega)=\frac{\mathrm{d} \theta}{\mathrm{dt}}=2 \mathrm{at}\) Angular acceleration \((\alpha)=\frac{\mathrm{d} \omega}{\mathrm{dt}}=2 \mathrm{a}\) Tangential linear acceleration, \(a_{\mathrm{t}} =\alpha \mathrm{R}\) \(\mathrm{a} =2 \mathrm{aR}\) Radial or normal acceleration of particle \(a_{r} =\omega^{2} R\) \(=(2 a t)^{2} R\) \(a_{r} =4 a^{2} t^{2} R\) From equation (i) and (ii) \(\mathrm{a}_{\text {total }} =\sqrt{\mathrm{a}_{\mathrm{t}}^{2}+\mathrm{a}_{\mathrm{r}}^{2}}=\sqrt{(2 \mathrm{aR})^{2}+\left(4 \mathrm{a}^{2} \mathrm{t}^{2} \mathrm{R}\right)^{2}}\) \(\mathrm{a}_{\text {total }} =2 \mathrm{aR} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}} \quad\left[\because 2 \mathrm{aR}=\frac{\mathrm{v}}{\mathrm{t}}\right]\) \(\therefore \quad \mathrm{a}_{\text {total }} =\frac{\mathrm{v}}{\mathrm{t}} \sqrt{1+4 \mathrm{a}^{2} \mathrm{t}^{4}}\)
TS- EAMCET-05.05.2018
Rotational Motion
149826
The angular speed of second's hand of the clock is
1 \(\frac{\pi}{1800} \operatorname{rad~s}^{-1}\)
2 \(\frac{\pi}{30} \operatorname{rad~s}^{-1}\)
3 \(\frac{\pi}{90} \mathrm{rads}^{-1}\)
4 \(\frac{\pi}{60} \mathrm{rads}^{-1}\)
Explanation:
B Given, Speed of second hand of the clock \(\mathrm{t}=60 \mathrm{sec}\). \(\theta=2 \pi \mathrm{rad}\) Angular speed \((\omega)=\frac{\text { Angular displacement }(\theta)}{\operatorname{time}(\mathrm{t})}\) \(=\frac{2 \pi}{60}\) \(\omega=\frac{\pi}{30} \mathrm{rad} / \mathrm{sec}\)
