Q. 72

Expert-verifiedFound in: Page 334

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

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?

a) Cylinder’s initial angular acceleration is 21.7 rad/sec^{2}

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

The mass of the cylinder = 5kg

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

Lets draw the FBD as below

Angular acceleration is calculated by

Torque can be calculated by

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

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

I_{CM} is the moment of inertia about it center of mass

Substitute in equation(4) we get

Substitute values in equation (1) to get angular acceleration

Substitute given values in the equation (5) we get

The mass of the cylinder = 5kg

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

To calculate the angular speed use the formula below

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/s^{2}

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