Informative line

### Angular Acceleration

Learn angular acceleration of a pulley and Relation between Torque and Angular Acceleration. Practice free body diagrams Atwood machine problems. Find its angular acceleration at the moment it is released.

# Idea of Fixed Axis

• For a rotating rigid body, the angular velocity $$\omega$$ and angular acceleration $$(\alpha)$$ of all the particles of the body are same.

• If the body is rigid i.e., distance between two points A and B doesn't change, then the angle rotated by A about O is same as that rotated by B.

Thus, $$\omega _A = \omega_B = \omega$$

Also, $$\alpha_A = \alpha_B = \alpha$$

## Relation between linear and rotational variables

• Consider a body and its particles A and B, as shown in figure.

• The displacement of A with respect to O is perpendicular to $$r_{A/O}$$ (position of A with respect to O) and is given by

$$ds = r_{A/O} \,d \,\theta$$.........(1)

• Dividing equation (1) by dt

$$\dfrac{ds}{dt} = r_{A/O} \dfrac{d\,\theta}{dt}$$

$$\Rightarrow v_{A/O} = r_{A/O} \,\omega$$  .......(2)

The velocity of A is perpendicular to

$${\vec r_{A/O}}$$ as well as $$\vec \omega$$

• Differentiating equation (2) with respect to time

$$\dfrac{dv_{A/O}}{dt} = r _{A/O} \dfrac{d\omega}{dt}$$

$$\Rightarrow a_{A/O} = r_{A/O} \,\alpha$$

#### A pulley is similar to a disk of radius R, rotating about a perpendicular axis passing through its center of mass with an angular acceleration $$\alpha$$. Two blocks attached by a string passing over the pulley are shown. If there is no slipping between the string and pulley, find the acceleration of the blocks in terms of R and $$\alpha$$.

A $$a_P = R \,\alpha$$ (downward) $$a_Q = R \,\alpha$$ (upward)

B $$a_P = 2R\,\alpha$$ (downward) $$a_Q = R\,\alpha$$ (upward)

C $$a_P = 2R\, \alpha$$ (downward) $$a_Q = 2R\, \alpha$$  (upward)

D $$a_P = 3R\,\alpha$$ (downward) $$a_Q = 2R\,\alpha$$ (upward)

×

Acceleration of block 1 is same as that of point P and that of block 2 is same as acceleration of point Q.

Acceleration of P with respect to O,

$$a_{P/O} = r_{P/O} \,\alpha$$

as $$\vec a _O = 0$$

Acceleration of P = position vector of P with respect to O × $$\alpha$$

$$a_{P} = r_{P/O} \ \alpha$$

$$=R\ \alpha$$  downward

Acceleration of Q =  position vector of Q with respect to O × $$\alpha$$

$$a_Q=r_{Q/O}\,\alpha$$

$$=R\ \alpha$$  upward

### A pulley is similar to a disk of radius R, rotating about a perpendicular axis passing through its center of mass with an angular acceleration $$\alpha$$. Two blocks attached by a string passing over the pulley are shown. If there is no slipping between the string and pulley, find the acceleration of the blocks in terms of R and $$\alpha$$.

A

$$a_P = R \,\alpha$$ (downward)

$$a_Q = R \,\alpha$$ (upward)

.

B

$$a_P = 2R\,\alpha$$ (downward)

$$a_Q = R\,\alpha$$ (upward)

C

$$a_P = 2R\, \alpha$$ (downward)

$$a_Q = 2R\, \alpha$$  (upward)

D

$$a_P = 3R\,\alpha$$ (downward)

$$a_Q = 2R\,\alpha$$ (upward)

Option A is Correct

# Relation between Torque and Angular Acceleration

• Consider a body of mass m, rotating about a fixed axis O.
• Torque applied on the body is given by -

$$\vec \tau_0 = \vec r × \vec F$$

$$\vec \tau_0 = \vec r × (m\,\vec a)$$

$$\vec \tau_0 = \vec r × (m\, r\, \alpha)$$               [  $$\because a = r \,\alpha$$]

$$\vec \tau_0 = (m\, r^2\,) \alpha$$

$$\vec \tau_0 = (I_0) \,\alpha$$      [$$\because I_0 = mr^2$$]

???$$\alpha = \dfrac{\tau_{axis}}{I_{axis}}$$

Hence, angular acceleration of a body is equal to the ratio of torque acting on the body about the axis to the moment of inertia of the body about that axis.

#### A rod of mass m and length $$\ell$$, hinged at one end is released from horizontal position, as shown. Find its angular acceleration at the moment it is released.

A $$\dfrac{3g}{2\ell}$$

B $$\dfrac{4g}{\ell}$$

C $$\dfrac{2g}{\ell}$$

D $$\dfrac{g}{\ell}$$

×

Free body diagram of a rod of mass m and length $$\ell$$

Torque is given by,

$$\vec \tau = \vec r × \vec F$$

$$\tau = r_\bot \ F$$

= $$\dfrac{\ell}{2} mg$$  (clockwise)

Angular acceleration can be calculated by

$$\tau_{axis} =I_{axis} \, \alpha$$

$$\alpha = \dfrac{\dfrac{\ell}{2} mg}{\dfrac{m\ell^2}{3}} = \dfrac{3g}{2\ell}$$

### A rod of mass m and length $$\ell$$, hinged at one end is released from horizontal position, as shown. Find its angular acceleration at the moment it is released.

A

$$\dfrac{3g}{2\ell}$$

.

B

$$\dfrac{4g}{\ell}$$

C

$$\dfrac{2g}{\ell}$$

D

$$\dfrac{g}{\ell}$$

Option A is Correct

#### A rod AB of mass m and length $$\ell$$ is hinged at a point O at a distance  $$\ell_1$$ from an end A. If it is released from a position shown, then find the angular acceleration of the rod just after the release.

A $$\ell g \,cos\theta$$

B $$\dfrac{\ell}{2} g \,cos\theta$$

C $$\dfrac{\left(\dfrac{\ell}{2} - \ell_1\right)g \,cos\theta}{\dfrac{\ell^2}{12}+ \left(\dfrac{\ell}{2}-\ell_1\right)^2}$$

D $$\ell\,g$$

×

Free body diagram of a rod of mass m and length $$\ell$$,

Angular acceleration can be calculated by

$$\tau_{axis} = I_{axis} \cdot \alpha$$

Taking $$\tau$$ about point O  (Hinge)

$$mg \, r _\bot = I _{axis}\cdot\alpha$$

$$mg \left(\dfrac{\ell}{2}-\ell_1\right) cos \theta = \left\{\dfrac{m\ell^2}{12} + m \left(\dfrac{\ell}{2} - \ell_1\right)^2 \right\} \alpha$$

$$\alpha = \dfrac{\left(\dfrac{\ell}{2}-\ell_1\right)g \,cos\theta}{\dfrac{\ell^2}{12}+ \left(\dfrac{\ell}{2}-\ell_1\right)^2}$$

### A rod AB of mass m and length $$\ell$$ is hinged at a point O at a distance  $$\ell_1$$ from an end A. If it is released from a position shown, then find the angular acceleration of the rod just after the release.

A

$$\ell g \,cos\theta$$

.

B

$$\dfrac{\ell}{2} g \,cos\theta$$

C

$$\dfrac{\left(\dfrac{\ell}{2} - \ell_1\right)g \,cos\theta}{\dfrac{\ell^2}{12}+ \left(\dfrac{\ell}{2}-\ell_1\right)^2}$$

D

$$\ell\,g$$

Option C is Correct

#### A disk shaped pulley of mass M, radius R is shown in figure. A block of mass m is attached to it with the help of a string coiled around the pulley. Find angular acceleration $$\alpha$$ of the pulley and acceleration of block.

A $$\dfrac{mg}{\left(m+ \dfrac{M}{2}\right)}$$

B $$mg$$

C $$\dfrac{M}{2}$$

D $$M$$

×

Free body diagram for a pulley of mass M and radius R,

Angular acceleration can be calculated by

$$\tau _{axis} = I _{axis} \cdot \alpha$$

$$T_1 R = \dfrac{MR^2}{2} \alpha$$

$$T_1 = \dfrac{MR \,\alpha}{2}$$ .......(1)

External force is given by

$$\vec F_{ext} = m \,\vec a_{cm}$$

For block : mg - T1 = ma      ..............(2)

For pulley : T - Mg - T1 = 0    ..........(3)

Acceleration of block is equal to the acceleration of point P of the pulley,

$$a = R\, \alpha$$                .......(4)

Substituting the values of T1 and a from equation (1) and (4) in equation (2),

$$mg - \dfrac{MR\,\alpha}{2} = mR \,\alpha$$

$$\alpha = \dfrac{mg}{\left(m+\dfrac{M}{2}\right)R}$$

From equation (4)

$$a = \dfrac{mg}{\left(m+\dfrac{M}{2}\right)}$$

### A disk shaped pulley of mass M, radius R is shown in figure. A block of mass m is attached to it with the help of a string coiled around the pulley. Find angular acceleration $$\alpha$$ of the pulley and acceleration of block.

A

$$\dfrac{mg}{\left(m+ \dfrac{M}{2}\right)}$$

.

B

$$mg$$

C

$$\dfrac{M}{2}$$

D

$$M$$

Option A is Correct

# Angular Acceleration of a Pulley (when two different forces applied on it)

• Consider a body of mass m, rotating about a fixed axis O.
• Torque applied on the body is given by -

$$\vec \tau_0 = \vec r × \vec F$$

$$\vec \tau_0 = \vec r × (m\vec a)$$

$$\vec \tau_0 = \vec r × (m\, r\, \alpha)$$       [ $$\because a = r \,\alpha$$]

$$\vec \tau_0 = (m\, r^2\,) \alpha$$

$$\vec \tau_0 = (I_0) \,\alpha$$             [ $$\because I_0 = mr^2$$]

$$\alpha = \dfrac{\tau_{axis}}{I_{axis}}$$

• Hence, angular acceleration of a body is equal to the ratio of torque acting on the body about the axis to the moment of inertia of the body about that axis.

#### A pulley of mass M and radius R has radius of gyration as K about its center of mass. If two forces F1 and F2 act on the pulley as shown, then find the angular acceleration of the pulley.

A $$\dfrac{F_1 R}{MK^2}$$

B $$\dfrac{F_2 R}{MK^2}$$

C $$\dfrac{(F_1 - F_2) R}{MK^2}$$

D $$\dfrac{(F_1 + F_2) R}{MK^2}$$

×

Free body diagram of a pulley of mass M and radius R

Angular acceleration can be calculated by

$$\tau_{axis} = I _{axis} \,\alpha$$

$$F_1 R - F_2 R = MK^2 \alpha$$

$$\alpha = \dfrac{(F_1 - F_2)R}{MK^2}$$

### A pulley of mass M and radius R has radius of gyration as K about its center of mass. If two forces F1 and F2 act on the pulley as shown, then find the angular acceleration of the pulley.

A

$$\dfrac{F_1 R}{MK^2}$$

.

B

$$\dfrac{F_2 R}{MK^2}$$

C

$$\dfrac{(F_1 - F_2) R}{MK^2}$$

D

$$\dfrac{(F_1 + F_2) R}{MK^2}$$

Option C is Correct

# Free Body Diagrams of Atwood Machine

•  Consider a pulley of mass M and radius R.
• The pulley has radius of gyration K about center of mass.
• An ideal string passes over the pulley with masses m1 and m2 attached at its end and as shown in figure.
• This system is known as Atwood machine.

• To calculate torque and thus, angular acceleration, shown free body diagram is used ( with all force at the original point of application ).
• Hence T1 and T2 are the tensions in the ideal string passes over the pulley with masses m1 and m2.

• To calculate $$\vec F_{net}$$ on the pulley and  thus acceleration of the center of mass of pulley shown free body diagram is used.
• Since, force at the same point is more convenient as there is no any effect of point of application of force on $$\vec F_{net}$$ and $$\vec a_{CM}$$.

• To show the forces acting on the blocks shown free body diagram is used.

#### A pulley of mass M and radius R has radius of gyration K about center of mass. An ideal string passes over the pulley with masses m1 and m2 attached at its ends, as shown in figure. Which one of the following free body diagrams is correct?

A

B

C

D

×

The figure shown, body diagram is used to calculate torque and thus angular acceleration. (with all the forces at their original point of application)

This free body diagram is used to calculate $$\vec F_{net}$$ on the pulley and thus acceleration of center of mass of pulley. Since, forces at the same point is more convenient as there is no any effect of point of application of force on $$\vec F_{net}$$ and $$\vec a_{cm}$$.

This free body diagram is used for the blocks.

Hence, option A is correct.

### A pulley of mass M and radius R has radius of gyration K about center of mass. An ideal string passes over the pulley with masses m1 and m2 attached at its ends, as shown in figure. Which one of the following free body diagrams is correct?

A
B
C
D

Option A is Correct

# Angular Acceleration of Pulley

• Consider a body of mass m, rotating about a fixed axis O.
• Torque applied on the body is given by

$$\vec \tau_0 = \vec r × \vec F$$

$$\vec \tau_0 = \vec r × (m\vec a)$$

$$\vec \tau_0 = \vec r × (m\, r\, \alpha)$$       [  $$\because a = r \,\alpha$$]

$$\vec \tau_0 = (m\, r^2\,) \alpha$$

$$\vec \tau_0 = (I_0) \,\alpha$$            [$$\because I_0 = mr^2$$]

$$\alpha = \dfrac{\tau_{axis}}{I_{axis}}$$

Hence, angular acceleration of a body is equal to the ratio of torque acting on the body about the axis to the moment of inertia of the body about that axis.

#### A pulley of mass M and radius R has radius of gyration K about its center of mass. An ideal string passes over the pulley with masses m1 and m2 attached to its ends, as shown in figure. If there is no slipping, then calculate the angular acceleration of the pulley.

A $$\dfrac{m_1 g}{\left(\dfrac{MK^2}{R^2}\right)R}$$

B $$\dfrac{(m_1-m_2) g}{\left(\dfrac{MK^2}{R^2}+m_1+m_2\right)R}$$

C $$\dfrac{m_1\,g}{R}$$

D $$\dfrac{MK^2}{g} R$$

×

Free body diagram for the pulley of mass M and radius R,

External force is given by,

$$\vec F _{ext} = m\, \vec a$$

$$\Rightarrow m_1g-T_1=m_1a$$     ....(1)

$$\Rightarrow T_2-m_2g=m_2a$$      .....(2)

$$\tau _{axis} = I_ {axis} \,\alpha$$

$$T_1 R - T_2 R = MK^2 \,\alpha$$

$$\alpha = \dfrac{R}{MK^2} (T_1 - T_2)$$  ......(3)

Acceleration of a point on a rigid body at a distance $$r_\bot$$ from the axis of rotation is given by

$$a = r_\bot \,\alpha$$

As there is no slipping, so $$\vec a$$ of block 1 is equal to $$\vec a$$ of point P and $$\vec a$$ of block 2 is equal to $$\vec a$$ of point Q.

$$\therefore a = R \,\alpha$$  ..... (4)

Putting the value of T1, T2 and $$a$$ from equation (1), (2) and (4) in equation (3) -

$$\alpha = \dfrac{R}{ MK^2} \{{m_1g - m_1 a - (m_2 a + m_2 g)}\}$$

$$\alpha = \dfrac{R}{ MK^2} \{{(m_1 - m_2 ) g - (m_1 + m_2)R \,\alpha}\}$$

$$\alpha = \dfrac{(m_1 -m_2)g}{\left(\dfrac{MK^2}{R^2}+ m_1+m_2\right) R}$$

### A pulley of mass M and radius R has radius of gyration K about its center of mass. An ideal string passes over the pulley with masses m1 and m2 attached to its ends, as shown in figure. If there is no slipping, then calculate the angular acceleration of the pulley.

A

$$\dfrac{m_1 g}{\left(\dfrac{MK^2}{R^2}\right)R}$$

.

B

$$\dfrac{(m_1-m_2) g}{\left(\dfrac{MK^2}{R^2}+m_1+m_2\right)R}$$

C

$$\dfrac{m_1\,g}{R}$$

D

$$\dfrac{MK^2}{g} R$$

Option B is Correct

#### In the figure shown, the pulley is a disk of mass M and radius R by which blocks of masses m1 and m2 are attached. Find the angular acceleration of pulley when the system is released from rest.

A $$m_1g$$

B $$\dfrac{m_1\,g}{\left(m_1+m_2 + \dfrac{M}{2}\right)R}$$

C $$m_2\,g$$

D $$M\,g \cdot R$$

×

Free body diagram of a pulley of mass M and radius R,

Angular acceleration can be calculated by

$$\tau_{axis} = I_{axis} \,\alpha$$

Taking $$\tau$$ about O

$$T_1 R - T_2 R = \dfrac{MR^2}{2}\, \alpha$$

$$T_1 - T_2 = \dfrac{MR\,\alpha}{2}\,$$...........(1)

External force is given by

$$\vec F_{ext} = m\,\vec a_{cm}$$

$$m_1 \,g - T_1 = m_1\,a$$       .......(2)

$$T_2 = m_2 \,a$$  ........(3)

Tension for pulley is,

$$N_y = T_1 + Mg$$           $$\,(F_y = 0)$$

$$T_2 = N_x$$                      $$(F_x = 0)$$

Acceleration of a point on a rigid body at a distance $$r_\bot$$ from the axis of rotation is given by

$$a = r_\bot . \alpha$$

As there no slipping, so $$\vec a$$ of block 1 is equal to $$\vec a$$ of point P and $$\vec a$$ of block 2 is equal to $$\vec a$$ of point Q.

$$\therefore a = R\, \alpha$$                  ............(4)

Adding equation(1), (2), (3) and substituting the value of $$a$$ from equation(4),

$$m_1 g = \left(\dfrac{M}{2} + m_1 + m_2\right) R \, \alpha$$

$$\alpha = \dfrac{m_1 g}{\left(m_1 + m_2 + \dfrac{M}{2}\right) R}$$

### In the figure shown, the pulley is a disk of mass M and radius R by which blocks of masses m1 and m2 are attached. Find the angular acceleration of pulley when the system is released from rest.

A

$$m_1g$$

.

B

$$\dfrac{m_1\,g}{\left(m_1+m_2 + \dfrac{M}{2}\right)R}$$

C

$$m_2\,g$$

D

$$M\,g \cdot R$$

Option B is Correct