Question

A solid cylindrical disk whose mass is 1.4 kg and radius is 8.5 cm, rolls across a horizontal table at a speed of 15 cm/...


A solid cylindrical disk whose mass is 1.4 kg and radius is 8.5 cm, rolls across a horizontal table at a speed of 15 cm/s. What is the instantaneous velocity of the top of the rolling cylinder? What is the angular speed of the rolling disk? What is the kinetic energy of the rolling disk?

Answer

Given:

The mass (m) of the solid cylindrical disk is 1.4 Kg.

The radius (r) of the disk is 8.5 cm.

The speed (v) of the rolling disk is 15 cm/s.

Introduction:

The moment of inertia (I) plays the same role in rotational mechanics as the mass plays in kinematics or in linear motion. Angular speed (ω) is defined as the time rate of change of angular displacement of the body. The kinetic energy of a body in rotational motion is the function of its moment of inertia and angular speed.

Calculation (instantaneous speed):

The given solid cylindrical disk is rolling in the horizontal direction at a speed of 15 cm/s. It means that at any particular instant of time and at any point on the surface of the disk, the speed of the disk is 15 cm/s.

Thus, the instantaneous speed at the top of the rolling cylinder is 15 cm/s. 

Calculation (angular speed):

Write the relationship between the linear speed and the angular speed of a rolling cylinder.

v=rωω=vr

Substitute, 15 cm/s for the value of v, and 8.5 cm for the value of r in the above expression.

ω=15 cm/s8.5 cmω=1.764 s-1

Thus, the angular speed of the rolling disk is 1.764 s-1.

Calculation (Rotational K.E.):

Write the expression for the moment of inertia for a cylinder about an axis through its center.

I=12mr2

Substitute, 1.4 Kg for the value of m, and 8.5 cm for the value of r in the above expression.

I=12(1.4 Kg)8.5 cm0.01 m1 cm2I=(0.5)(1.4 Kg)(0.085 m)2 I=0.0050575 Kg.m2

Write the expression for the rotational kinetic energy.

Erot=12Iω2

Substitute, 0.0050575 Kg.m2 for the value of I, and 1.764 s-1 for the value of ω in the above expression.

Erot=12(0.0050575 Kg.m2)(1.764 s-1)2 Erot=(0.5)(0.0050575 Kg.m2)(3.111696 s-2)Erot=0.007868 J

Thus, the rotational kinetic energy of the given rolling disk is 0.007868 J.

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