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Rotational Motion

149827 A solid body rotates an angle $θ$ about a stationary axis according to the law $θ = 6 t - 2 t^{3}$ . What is the mean value of angular velocity over the time interval between $t = 0$ and the time when the body comes to rest?

1

1 rad / s

2

2 rad / s

3

3 rad / s

4

4 rad / s

Explanation:

D Given that,
$θ = 6 t - 2 t^{3}$
Differentiating with respect to $t$ on both side of the equation-
$\frac{d θ}{dt} = ω = 6 - 6 t^{2}$
$6 - 6 t^{2} = 0$
$t^{2} = 1$
$t = 1 \sec$
Put the value of $t$ in equation (i) we get -
$θ = 6 \times 1 - 2 \times 1^{3}$
$θ = 6 - 2$
$θ = 4 rad$
$For θ_{0} = 6 \times 0 - 2 \times 0^{2} [∵ t_{0} = 0]$
$θ_{0} = 0 rad$
Mean value of angular velocity $ω = \frac{θ - θ_{0}}{t - t_{0}} = \frac{4 - 0}{1 - 0}$
$ω = 4 rad / \sec$

UPSEE - 2018

Rotational Motion

149828 A wheel rotating at $12 rev / s$ is brought to rest in $6 s$ . The average angular deceleration in $rad / s^{2}$ of the wheel during this process is

1

4 π

2 4

3 72

4

1 / π

5

π

Explanation:

A Given ;
$ω_{i} = 2 π N$
$ω_{i} = 2 π \times 12 rad / \sec$
$ω_{i} = 24 π rad / \sec$
$t = 6 \sec, ω_{f} = 0$
From,
$ω_{f} = ω_{i} - α Δ t$
$α = \frac{ω_{i}}{Δ t}$
$= \frac{24 π}{6}$
$α = 4 π rad / \sec^{2}$

Kerala CEE - 2017

Rotational Motion

149829 A uniform disc of mass $M$ and radius $R$ has a string wound around its periphery and tied to a ceiling as shown in the figure. The acceleration of the center of mass after it is released

1

g

2

\frac{g}{3}

3

\frac{2 g}{3}

4

\frac{g}{2}

Explanation:

C
Let us consider
Equation of motion of the disc
$Ma = Mg - T$
Also torque about $O$ .
$TR = I α$
$∴ TR = (\frac{1}{2} {MR}^{2}) α$
For no slipping, $a = α R \Rightarrow α = \frac{a}{R}$
We get, $T = \frac{Ma}{2}$
From equation (i) and (ii), $Ma = Mg - \frac{Ma}{2}$
$ma = m (g - \frac{a}{2})$
$a + \frac{a}{2} = g$
$\frac{2 a + a}{2} = g$
$a = \frac{2 g}{3}$

TS EAMCET(Medical)-2017

Rotational Motion

149830 A rod $P Q$ of mass $M$ and length $L$ is hinged at end $P$ . The rod is kept horizontal by a massless string tied to point $O$ as shown in figure.
original image
When string is cut, the initial angular acceleration of the rod is-

1

\frac{3 g}{2 L}

2

\frac{g}{L}

3

\frac{2 g}{L}

4

\frac{2 g}{2 L}

Explanation:

A original image
$P ⟶ \underset{mg}{↓} Q$ (Free body diagram)
angular torque, $τ = I α$
Moment of inertia (I) about the point $Q$ on the rad,
$I = \frac{{mL}^{2}}{3}$
Torque due to weight,
$τ = mg (\frac{L}{2})$
So, putting the value of $τ$ and $I$ in $τ = I α$ -
$mg (\frac{L}{2}) = \frac{{mL}^{2}}{3} α$
$α = \frac{3}{2} \frac{g}{L}$
$α = \frac{3 g}{2 t}$

BCECE-2017

Rotational Motion

149827 A solid body rotates an angle $θ$ about a stationary axis according to the law $θ = 6 t - 2 t^{3}$ . What is the mean value of angular velocity over the time interval between $t = 0$ and the time when the body comes to rest?

1

1 rad / s

2

2 rad / s

3

3 rad / s

4

4 rad / s

Explanation:

D Given that,
$θ = 6 t - 2 t^{3}$
Differentiating with respect to $t$ on both side of the equation-
$\frac{d θ}{dt} = ω = 6 - 6 t^{2}$
$6 - 6 t^{2} = 0$
$t^{2} = 1$
$t = 1 \sec$
Put the value of $t$ in equation (i) we get -
$θ = 6 \times 1 - 2 \times 1^{3}$
$θ = 6 - 2$
$θ = 4 rad$
$For θ_{0} = 6 \times 0 - 2 \times 0^{2} [∵ t_{0} = 0]$
$θ_{0} = 0 rad$
Mean value of angular velocity $ω = \frac{θ - θ_{0}}{t - t_{0}} = \frac{4 - 0}{1 - 0}$
$ω = 4 rad / \sec$

UPSEE - 2018

Rotational Motion

149828 A wheel rotating at $12 rev / s$ is brought to rest in $6 s$ . The average angular deceleration in $rad / s^{2}$ of the wheel during this process is

1

4 π

2 4

3 72

4

1 / π

5

π

Explanation:

A Given ;
$ω_{i} = 2 π N$
$ω_{i} = 2 π \times 12 rad / \sec$
$ω_{i} = 24 π rad / \sec$
$t = 6 \sec, ω_{f} = 0$
From,
$ω_{f} = ω_{i} - α Δ t$
$α = \frac{ω_{i}}{Δ t}$
$= \frac{24 π}{6}$
$α = 4 π rad / \sec^{2}$

Kerala CEE - 2017

Rotational Motion

149829 A uniform disc of mass $M$ and radius $R$ has a string wound around its periphery and tied to a ceiling as shown in the figure. The acceleration of the center of mass after it is released

1

g

2

\frac{g}{3}

3

\frac{2 g}{3}

4

\frac{g}{2}

Explanation:

C
Let us consider
Equation of motion of the disc
$Ma = Mg - T$
Also torque about $O$ .
$TR = I α$
$∴ TR = (\frac{1}{2} {MR}^{2}) α$
For no slipping, $a = α R \Rightarrow α = \frac{a}{R}$
We get, $T = \frac{Ma}{2}$
From equation (i) and (ii), $Ma = Mg - \frac{Ma}{2}$
$ma = m (g - \frac{a}{2})$
$a + \frac{a}{2} = g$
$\frac{2 a + a}{2} = g$
$a = \frac{2 g}{3}$

TS EAMCET(Medical)-2017

Rotational Motion

149830 A rod $P Q$ of mass $M$ and length $L$ is hinged at end $P$ . The rod is kept horizontal by a massless string tied to point $O$ as shown in figure.
original image
When string is cut, the initial angular acceleration of the rod is-

1

\frac{3 g}{2 L}

2

\frac{g}{L}

3

\frac{2 g}{L}

4

\frac{2 g}{2 L}

Explanation:

A original image
$P ⟶ \underset{mg}{↓} Q$ (Free body diagram)
angular torque, $τ = I α$
Moment of inertia (I) about the point $Q$ on the rad,
$I = \frac{{mL}^{2}}{3}$
Torque due to weight,
$τ = mg (\frac{L}{2})$
So, putting the value of $τ$ and $I$ in $τ = I α$ -
$mg (\frac{L}{2}) = \frac{{mL}^{2}}{3} α$
$α = \frac{3}{2} \frac{g}{L}$
$α = \frac{3 g}{2 t}$

BCECE-2017

Rotational Motion

149827 A solid body rotates an angle $θ$ about a stationary axis according to the law $θ = 6 t - 2 t^{3}$ . What is the mean value of angular velocity over the time interval between $t = 0$ and the time when the body comes to rest?

1

1 rad / s

2

2 rad / s

3

3 rad / s

4

4 rad / s

Explanation:

D Given that,
$θ = 6 t - 2 t^{3}$
Differentiating with respect to $t$ on both side of the equation-
$\frac{d θ}{dt} = ω = 6 - 6 t^{2}$
$6 - 6 t^{2} = 0$
$t^{2} = 1$
$t = 1 \sec$
Put the value of $t$ in equation (i) we get -
$θ = 6 \times 1 - 2 \times 1^{3}$
$θ = 6 - 2$
$θ = 4 rad$
$For θ_{0} = 6 \times 0 - 2 \times 0^{2} [∵ t_{0} = 0]$
$θ_{0} = 0 rad$
Mean value of angular velocity $ω = \frac{θ - θ_{0}}{t - t_{0}} = \frac{4 - 0}{1 - 0}$
$ω = 4 rad / \sec$

UPSEE - 2018

Rotational Motion

149828 A wheel rotating at $12 rev / s$ is brought to rest in $6 s$ . The average angular deceleration in $rad / s^{2}$ of the wheel during this process is

1

4 π

2 4

3 72

4

1 / π

5

π

Explanation:

A Given ;
$ω_{i} = 2 π N$
$ω_{i} = 2 π \times 12 rad / \sec$
$ω_{i} = 24 π rad / \sec$
$t = 6 \sec, ω_{f} = 0$
From,
$ω_{f} = ω_{i} - α Δ t$
$α = \frac{ω_{i}}{Δ t}$
$= \frac{24 π}{6}$
$α = 4 π rad / \sec^{2}$

Kerala CEE - 2017

Rotational Motion

149829 A uniform disc of mass $M$ and radius $R$ has a string wound around its periphery and tied to a ceiling as shown in the figure. The acceleration of the center of mass after it is released

1

g

2

\frac{g}{3}

3

\frac{2 g}{3}

4

\frac{g}{2}

Explanation:

C
Let us consider
Equation of motion of the disc
$Ma = Mg - T$
Also torque about $O$ .
$TR = I α$
$∴ TR = (\frac{1}{2} {MR}^{2}) α$
For no slipping, $a = α R \Rightarrow α = \frac{a}{R}$
We get, $T = \frac{Ma}{2}$
From equation (i) and (ii), $Ma = Mg - \frac{Ma}{2}$
$ma = m (g - \frac{a}{2})$
$a + \frac{a}{2} = g$
$\frac{2 a + a}{2} = g$
$a = \frac{2 g}{3}$

TS EAMCET(Medical)-2017

Rotational Motion

149830 A rod $P Q$ of mass $M$ and length $L$ is hinged at end $P$ . The rod is kept horizontal by a massless string tied to point $O$ as shown in figure.
original image
When string is cut, the initial angular acceleration of the rod is-

1

\frac{3 g}{2 L}

2

\frac{g}{L}

3

\frac{2 g}{L}

4

\frac{2 g}{2 L}

Explanation:

A original image
$P ⟶ \underset{mg}{↓} Q$ (Free body diagram)
angular torque, $τ = I α$
Moment of inertia (I) about the point $Q$ on the rad,
$I = \frac{{mL}^{2}}{3}$
Torque due to weight,
$τ = mg (\frac{L}{2})$
So, putting the value of $τ$ and $I$ in $τ = I α$ -
$mg (\frac{L}{2}) = \frac{{mL}^{2}}{3} α$
$α = \frac{3}{2} \frac{g}{L}$
$α = \frac{3 g}{2 t}$

BCECE-2017

Rotational Motion

149827 A solid body rotates an angle $θ$ about a stationary axis according to the law $θ = 6 t - 2 t^{3}$ . What is the mean value of angular velocity over the time interval between $t = 0$ and the time when the body comes to rest?

1

1 rad / s

2

2 rad / s

3

3 rad / s

4

4 rad / s

Explanation:

D Given that,
$θ = 6 t - 2 t^{3}$
Differentiating with respect to $t$ on both side of the equation-
$\frac{d θ}{dt} = ω = 6 - 6 t^{2}$
$6 - 6 t^{2} = 0$
$t^{2} = 1$
$t = 1 \sec$
Put the value of $t$ in equation (i) we get -
$θ = 6 \times 1 - 2 \times 1^{3}$
$θ = 6 - 2$
$θ = 4 rad$
$For θ_{0} = 6 \times 0 - 2 \times 0^{2} [∵ t_{0} = 0]$
$θ_{0} = 0 rad$
Mean value of angular velocity $ω = \frac{θ - θ_{0}}{t - t_{0}} = \frac{4 - 0}{1 - 0}$
$ω = 4 rad / \sec$

UPSEE - 2018

Rotational Motion

149828 A wheel rotating at $12 rev / s$ is brought to rest in $6 s$ . The average angular deceleration in $rad / s^{2}$ of the wheel during this process is

1

4 π

2 4

3 72

4

1 / π

5

π

Explanation:

A Given ;
$ω_{i} = 2 π N$
$ω_{i} = 2 π \times 12 rad / \sec$
$ω_{i} = 24 π rad / \sec$
$t = 6 \sec, ω_{f} = 0$
From,
$ω_{f} = ω_{i} - α Δ t$
$α = \frac{ω_{i}}{Δ t}$
$= \frac{24 π}{6}$
$α = 4 π rad / \sec^{2}$

Kerala CEE - 2017

Rotational Motion

149829 A uniform disc of mass $M$ and radius $R$ has a string wound around its periphery and tied to a ceiling as shown in the figure. The acceleration of the center of mass after it is released

1

g

2

\frac{g}{3}

3

\frac{2 g}{3}

4

\frac{g}{2}

Explanation:

C
Let us consider
Equation of motion of the disc
$Ma = Mg - T$
Also torque about $O$ .
$TR = I α$
$∴ TR = (\frac{1}{2} {MR}^{2}) α$
For no slipping, $a = α R \Rightarrow α = \frac{a}{R}$
We get, $T = \frac{Ma}{2}$
From equation (i) and (ii), $Ma = Mg - \frac{Ma}{2}$
$ma = m (g - \frac{a}{2})$
$a + \frac{a}{2} = g$
$\frac{2 a + a}{2} = g$
$a = \frac{2 g}{3}$

TS EAMCET(Medical)-2017

Rotational Motion

149830 A rod $P Q$ of mass $M$ and length $L$ is hinged at end $P$ . The rod is kept horizontal by a massless string tied to point $O$ as shown in figure.
original image
When string is cut, the initial angular acceleration of the rod is-

1

\frac{3 g}{2 L}

2

\frac{g}{L}

3

\frac{2 g}{L}

4

\frac{2 g}{2 L}

Explanation:

A original image
$P ⟶ \underset{mg}{↓} Q$ (Free body diagram)
angular torque, $τ = I α$
Moment of inertia (I) about the point $Q$ on the rad,
$I = \frac{{mL}^{2}}{3}$
Torque due to weight,
$τ = mg (\frac{L}{2})$
So, putting the value of $τ$ and $I$ in $τ = I α$ -
$mg (\frac{L}{2}) = \frac{{mL}^{2}}{3} α$
$α = \frac{3}{2} \frac{g}{L}$
$α = \frac{3 g}{2 t}$

BCECE-2017

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