Rotational Motion — Page 5 | JEE Physics

Q41

A ball is spun with angular acceleration

\alpha

= 6t2

-

2t where t is in second and

\alpha

is in rads

-

2. At t = 0, the ball has angular velocity of 10 rads

-

1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :

A

{3 \over 2}{t^4} - {t^2} + 10t

B

{{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4

C

{{2{t^4}} \over 3} - {{{t^3}} \over 6} + 10t + 12

D

2{t^4} - {{{t^3}} \over 2} + 5t + 4

Correct Answer

Option B

Solution

\alpha = {{d\omega } \over {dt}} = 6{t^2} - 2t

\int_0^\omega {d\omega = \int_0^t {(6{t^2} - 2t)dt} }

so

\omega = 2{t^3} - {t^2} + 10

and

{{d\theta } \over {dt}} = 2{t^3} - {t^2} + 10

so

\int_4^\theta {d\theta = \int_0^t {(2{t^3} - {t^2} + 10)dt} }

\theta = {{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4

Q42

One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity

\omega

about an axis passing through fixed end, then the elongation of the spring will be :

A

{{k - m{\omega ^2}{l_0}} \over {m{\omega ^2}}}

B

{{m{\omega ^2}{l_0}} \over {k + m{\omega ^2}}}

C

{{m{\omega ^2}{l_0}} \over {k - m{\omega ^2}}}

D

{{k + m{\omega ^2}{l_0}} \over {m{\omega ^2}}}

Correct Answer

Option C

Solution

m{\omega ^2}({l_0} + x) = kx

\Rightarrow m{\omega ^2}{l_0} = (k - m{\omega ^2}) \times x

\Rightarrow x = {{m{\omega ^2}{l_0}} \over {(k - m{\omega ^2})}}

Q43

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads

-

1 in a horizontal plane about an axis vertical to its plane and passing through the center of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the ring will then rotate with an angular velocity (in rads

-

1).

A

{M \over {(M + m)}}

B

{{(M + 2m)} \over {2M}}

C

{{2M} \over {(M + 2m)}}

D

{{2(M + 2m)} \over M}

Correct Answer

Option C

Solution

{I_1}{\omega _1} = {I_2}{\omega _2}

M{R^2}{\omega _1} = (M{R^2} + 2m{R^2}){\omega _2}

{\omega _2} = \left( {{M \over {M + 2m}}} \right){\omega _1}

{\omega _2} = 2\left( {{M \over {M + 2m}}} \right)

Q44

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is

A

{2 \over 5}

B

{2 \over 7}

C

{1 \over 5}

D

{7 \over 10}

Correct Answer

Option B

Solution

K{E_R} = {1 \over 2}l{w^2}

= {1 \over 2} \times {2 \over 5} \times {\omega ^2} \times (m{R^2})

K{E_{total}} = {1 \over 2} \times {7 \over 5} \times m{R^2} \times {\omega ^2}

$\therefore$

{{K{E_R}} \over {K{E_{total}}}} = {2 \over 7}

Q45

A solid cylinder and a solid sphere, having same mass

M

and radius

R

, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be :

A

\sqrt{\dfrac{5}{3}}

B

\sqrt{\dfrac{4}{5}}

C

\sqrt{\dfrac{3}{5}}

D

\sqrt{\dfrac{14}{15}}

Correct Answer

Option D

Solution

a = {{g\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}

v = \sqrt {{{2Sg\sin \theta } \over {1 + {{{K^2}} \over {{R^2}}}}}}

\Rightarrow {{{v_c}} \over {{v_{ss}}}}\sqrt {{{1 + {{K_{ss}^2} \over {{R^2}}}} \over {1 + {{K_c^2} \over {{R^2}}}}}} = \sqrt {{{1 + {2 \over 5}} \over {1 + {1 \over 2}}}}

\Rightarrow \sqrt {{{{7 \over 5}} \over {{3 \over 2}}}} = \sqrt {{{14} \over {15}}}

Q46

A thin circular disc of mass

\mathrm{M}

and radius

\mathrm{R}

is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with angular velocity

\omega

. If another disc of same dimensions but of mass

\mathrm{M} / 2

is placed gently on the first disc co-axially, then the new angular velocity of the system is :

A

\dfrac{4}{5} \omega

B

\dfrac{5}{4} \omega

C

\dfrac{3}{2} \omega

D

\dfrac{2}{3} \omega

Correct Answer

Option D

Solution

To determine the new angular velocity of the system, we use the principle of conservation of angular momentum.

When no external torque acts on a system, its angular momentum remains constant.

Let's denote the initial angular momentum and the final angular momentum, respectively, as

L_{\text{initial}}

and

L_{\text{final}}

. The initial angular momentum of the system is given by:

L_{\text{initial}} = I_{\text{initial}} \cdot \omega

Here,

I_{\text{initial}}

is the moment of inertia of the first disc. For a thin circular disc, the moment of inertia about its center is:

I_{\text{initial}} = \frac{1}{2} M R^2

Thus,

L_{\text{initial}} = \left( \frac{1}{2} M R^2 \right) \omega

When the second disc is placed gently on the first disc, the two discs rotate together with a common angular velocity

\omega'

. The moment of inertia of the second disc is:

I_{\text{second}} = \frac{1}{2} \left( \frac{M}{2} \right) R^2 = \frac{1}{4} M R^2

The combined moment of inertia of the system after placing the second disc is:

I_{\text{final}} = I_{\text{initial}} + I_{\text{second}} = \frac{1}{2} M R^2 + \frac{1}{4} M R^2 = \frac{3}{4} M R^2

Thus, the final angular momentum of the system is:

L_{\text{final}} = I_{\text{final}} \cdot \omega' = \left( \frac{3}{4} M R^2 \right) \omega'

By the conservation of angular momentum:

L_{\text{initial}} = L_{\text{final}}

This simplifies to:

\left( \frac{1}{2} M R^2 \right) \omega = \left( \frac{3}{4} M R^2 \right) \omega'

Solving for

\omega'

:

\omega' = \frac{\left( \frac{1}{2} M R^2 \right) \omega}{\left( \frac{3}{4} M R^2 \right)} = \frac{\omega}{\frac{3}{2}} = \frac{2}{3} \omega

So, the new angular velocity of the system is:

\omega' = \frac{2}{3} \omega

Thus, the correct answer is: Option D:

\frac{2}{3} \omega

Q47

A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is

A 290 g

B 200 g

C 190 g

D 300 g

Correct Answer

Option C

Solution

\begin{aligned} & \tau_{\text{Net }}=0 \Rightarrow(400 \mathrm{~g} \times 30)=(250 \mathrm{~g} \times 10)(\mathrm{mg} \times 50) \\ & \mathrm{m}=\frac{12000-2500}{50}=\frac{9500}{50} \\ & \mathrm{M}=190 \mathrm{~g} \end{aligned}

Q48

A solid sphere of mass '

m

' and radius '

r

' is allowed to roll without slipping from the highest point of an inclined plane of length '

L

' and makes an angle

30^{\circ}

with the horizontal. The speed of the particle at the bottom of the plane is

v_1

. If the angle of inclination is increased to

45^{\circ}

while keeping

L

constant. Then the new speed of the sphere at the bottom of the plane is

v_2

. The ratio

v_1^2: v_2^2

is

A

1: 3

B

1: 2

C

1: \sqrt{2}

D

1: \sqrt{3}

Correct Answer

Option C

Solution

Let's analyze the problem step-by-step.

Energy conservation: When the sphere rolls without slipping, its gravitational potential energy converts into both translational and rotational kinetic energy.

The energy conservation equation is given by:

mgL\sin\theta = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

where: $L\sin\theta$ is the vertical height, $I$ is the moment of inertia, and $\omega$ is the angular speed.

Moment of Inertia and Rolling Condition: For a solid sphere, the moment of inertia about its center is:

I = \frac{2}{5}mr^2

Since the sphere rolls without slipping, the linear speed $v$ and the angular speed $\omega$ are related by:

v = r\omega \quad \Rightarrow \quad \omega = \frac{v}{r}

Total Kinetic Energy: Substituting the rolling condition into the rotational kinetic energy gives: $\begin{aligned} K &= \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right) \left(\dfrac{v}{r}\right)^2 \\ &= \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right) \dfrac{v^2}{r^2} \\ &= \dfrac{1}{2}mv^2 + \dfrac{1}{5}mv^2 \\ &= \dfrac{7}{10}mv^2 \end{aligned}$ Finding $v^2$ in Terms of $L$ and $\sin\theta$ : Equate the initial potential energy to the final kinetic energy:

mgL\sin\theta = \frac{7}{10}mv^2

Solving for $v^2$ :

v^2 = \frac{10}{7}gL\sin\theta

Apply to the Two Cases: For $\theta = 30^\circ$ , since $\sin(30^\circ) = \dfrac{1}{2}$ :

v_1^2 = \frac{10}{7}gL\left(\frac{1}{2}\right) = \frac{5}{7}gL

For $\theta = 45^\circ$ , since $\sin(45^\circ) = \dfrac{1}{\sqrt{2}}$ :

v_2^2 = \frac{10}{7}gL\left(\frac{1}{\sqrt{2}}\right)

Calculate the Ratio $\dfrac{v_1^2}{v_2^2}$ : $\begin{aligned} \dfrac{v_1^2}{v_2^2} &= \dfrac{\left(\dfrac{5}{7}gL\right)}{\left(\dfrac{10}{7}gL\dfrac{1}{\sqrt{2}}\right)} \\ &= \dfrac{5}{10}\sqrt{2} \\ &= \dfrac{\sqrt{2}}{2} \end{aligned}$ Expressing this ratio as $v_1^2 : v_2^2$ , we have:

v_1^2 : v_2^2 = 1 : \sqrt{2}

Conclusion: According to the options given, the correct answer is: Option C: $1:\sqrt{2}$ .

Q49

A circular disk of radius R meter and mass M kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that

\theta(t)=5 t^2-8 t

, where

\theta(t)

is the angular position of the rotating disc as a function of time

t

. How much power is delivered by the applied torque, when

t=2 \mathrm{~s}

?

A

60 \mathrm{MR}^2

B

72 \mathrm{MR}^2

C

8 \mathrm{MR}^2

D

108 \mathrm{MR}^2

Correct Answer

Option A

Solution

Moment of Inertia and Angular Motion For a solid circular disk, the moment of inertia is given by

I = \frac{1}{2}MR^2.

The angular position is defined as

\theta(t) = 5t^2 - 8t.

Angular Velocity and Acceleration Differentiate with respect to time to obtain the angular velocity:

\omega(t) = \frac{d\theta}{dt} = 10t - 8.

Differentiating again, the angular acceleration is:

\alpha(t) = \frac{d\omega}{dt} = 10.

At time

t = 2 \, \text{s}

: Angular velocity:

\omega(2) = 10(2) - 8 = 12 \, \text{rad/s}.

Angular acceleration:

\alpha(2) = 10 \, \text{rad/s}^2.

Torque Calculation The torque applied by the external force is related to the moment of inertia and angular acceleration:

\tau = I \alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2.

Power Delivered Power delivered by a torque is given by:

P = \tau \omega.

At

t = 2 \, \text{s}

, substitute the values:

P = 5MR^2 \times 12 = 60MR^2.

Thus, the power delivered by the applied torque at

t = 2 \, \text{s}

is:

\boxed{60MR^2}.

Q50

Moment of inertia of a rod of mass ' M ' and length ' L ' about an axis passing through its center and normal to its length is '

\alpha

'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is :

A

\alpha / 4

B

\alpha / 8

C

\alpha

D

\alpha / 2

Correct Answer

Option A

Solution

\alpha=\frac{\mathrm{M} \ell^2}{12}\quad\text{..... (1)}

\begin{aligned} &\begin{aligned} & \alpha^{\prime}=2\left[\frac{\frac{\mathrm{M}}{2}\left(\frac{\ell}{2}\right)^2}{12}\right] \\ & \alpha^{\prime}=\frac{\mathrm{M} \ell^2}{48}=\frac{\alpha}{4} \end{aligned}\\ &\text{ Correct option is (2) } \end{aligned}