Question

A 3.56 kg block resting on a smooch horizontal surface is attached to a string that passes over a frictionless pulley as shown in the figure. The pulley has a radius of 0.28 m. A force of 7.5 N pulls .the other end of the string. As a result, the block moves a distance of 0.25 m starting from rest. Find the (a) acceleration of the block, (b) angular acceleration of the

pulley, and (c) moment of inertia of the pulley.

Answer 1

(a) Acceleration of the block
`a = \(2.10 \, \text{m/s}^2\)`

(b) Angular acceleration of the pulley
`(\(\alpha\)) = \(7.50 \, \text{rad/s}^2\)`

(c) Moment of inertia of the pulley
`(\(I\)) = \(0.28 \, \text{kg} \cdot \text{m}^2\)`

(a) Acceleration of the Block:

Using Newton's second law (F = ma), we can find the acceleration (a) of the block.

`$\[ a = \frac{7.5 \, \text{N}}{3.56 \, \text{kg}} \]`

`\[ a \approx 2.10 \, \text{m/s}^2 \]`

(b) Angular Acceleration of the Pulley:

`$\[ \alpha = (a)/(r) \]`

`$\[ \alpha = \frac{2.10 \, \text{m/s}^2}{0.28 \, \text{m}} \]`

`\[ \alpha \approx 7.50 \, \text{rad/s}^2 \]`

(c) Moment of Inertia of the Pulley:

`\[ \tau = Fr \]`

`\[ \tau = (7.5 \, \text{N})(0.28 \, \text{m}) \]`

`\[ \tau \approx 2.10 \, \text{Nm} \]`

Now, use the torque to find the moment of inertia:

`\[ I = (\tau)/(\alpha) \]`

`$\[ I = \frac{2.10 \, \text{Nm}}{7.50 \, \text{rad/s}^2} \]`

`\[ I \approx 0.28 \, \text{kg} \cdot \text{m}^2 \]`

So, the calculated values are:

(a) Acceleration of the block
`a = \(2.10 \, \text{m/s}^2\)`

(b) Angular acceleration of the pulley
`(\(\alpha\)) = \(7.50 \, \text{rad/s}^2\)`

(c) Moment of inertia of the pulley
`(\(I\)) = \(0.28 \, \text{kg} \cdot \text{m}^2\)`