Question

The 5.0 kg, 60-cm-diameter disk in FIGURE P12.72 rotates on an axle passing
through one edge. The axle is parallel to the floor. The cylinder is held with the
center of mass at the same height as the axle, then released.
a. What is the cylinder’s initial angular acceleration?
b. What is the cylinder’s angular velocity when it is directly below the axle?

Short answer

a) Cylinder’s initial angular acceleration is 21.7 rad/sec²

b) Cylinder’s angular velocity when it is directly below the axle is 8.25 rad/sec

Step-by-step solution

Part (a) Step 1: Given information

The mass of the cylinder = 5kg
The diameter of the cylinder = 60 cm
So radius = 30 cm = 0.3 m

Part(a) Step2: Explanation

Lets draw the FBD as below

Angular acceleration is calculated by

$\alpha =\frac{\tau }{I}.............................\left(1\right)$

Torque can be calculated by

$\tau =F\times l................................\left(2\right)$

Where, F is the net force acting on the cylinder and l is the length of arm.

Only force acting is gravitational force

Substitute in equation(2) G = mg

$\tau =mgR......................\left(3\right)$

This is in the shape of disc so use the moment of inertia of disc.

$I=I_{C}M+M{R}^2......................\left(4\right)$

I_CM is the moment of inertia about it center of mass $\frac{M{R}^2}{2}$

Substitute in equation(4) we get

$I=\frac{3M{R}^2}{3}$

Substitute values in equation (1) to get angular acceleration

$$\begin{gathered} \alpha =\frac{MgR}{\frac{3}{2}M{R}^2} \\ \alpha =\frac{2g}{3R}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(5\right) \end{gathered}$$

Substitute given values in the equation (5) we get

$$\begin{gathered} \alpha =\frac{2\times 9.8{\mathrm{ms}}^{-2}}{3\times 0.3\mathrm{m}} \\ =21.8{\mathrm{rads}}^{-2} \end{gathered}$$

Part(b) Step 1 : Given information

The mass of the cylinder = 5kg
The diameter of the cylinder = 60 cm
So radius = 30 cm = 0.3 m

Part(b) Step 2 : Explanation

To calculate the angular speed use the formula below

${\omega }_f^2-{\omega }_i^2=2\alpha \theta ,\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(6\right)$

Where

ω_f= final velocity, ω_i = initial velocity ,θ is angular displacement and α is angular displacement.

Substitute values: s the cylinder is below axle so angular displacement ,θ =π/2

ω_i =0 and α =21.8 rad/s²

$$\begin{gathered} {\omega }_f^2-(0{)}^2=2\times (21.8rad/s^2)\times (\pi /2) \\ {\omega }_f=8.25rad/s \end{gathered}$$