Question

A cylinder is 0.10 m in radius and 0.20 in length. Its rotational inertia, about the cylinder axis on which it is mounted, is 0.020 kg  m2. A string is wound around the cylinder and pulled with a force of 1.0 N. The angular acceleration of the cylinder is:

Answer 1

Answer:$5 rad/s^2$

Explanation:

Given

Radius of cylinder r=0.1 m

Length L=0.2 in.

Moment of inertia I=0.020 kg-m^2

Force F=1 N

We Know Torque is given by

$Torque =I\alpha =F\cdot r$

where $\alpha =angular\ acceleration$

$I\alpha =F\cdot r$

$0.02\cdot \alpha =1\cdot 0.1$

$\alpha =5 rad/s^2$    

Answer 2

The angular acceleration of the cylinder is 5 rad/s².

To calculate the angular acceleration of the cylinder, we use the formula of torque below.

What is torque?

Torque is the force that causes a body to rotate about an axis.

Formula:

• Iα = Fr........... Equation 1

Where:

• I = rotational initial
• α = Angular acceleration
• r = Radius of the cylinder
• F = Force
• θ = Angle.

Make α the subject of the equation

• α  = RFsinθ/I............. Equation 2

From the question,

Given:

• r = 0.1 m
• F = 1.0 N
• I = 0.020 kgm²
• θ = 90° ( Since the rope is tangental to the side of the cylinder).

Substitute these values into equation 2

• α = (1.0×0.1×sin90°)/(0.02)
• α = 5 rad/s²

Hence, The angular acceleration of the cylinder is 5 rad/s².

Learn more about torque here: https://brainly.com/question/20691242