Question

A 11.8-kg cylinder rolls without slipping on a rough surface. At an instant when its center of gravity has a speed of 10.4 m/s, determine the following. (a) the translational kinetic energy of its center of gravity J (b) the rotational kinetic energy about its center of gravity J (c) its total kinetic energy J

Answer 1

1. 638.144j

2. 319.072 j

3. 957.216 j

Step-by-step explanation:

M = 11.8 kg cylinder

V = speed of center of gravity = 10.4m/s

A.

Translational kinetic energy

= 1/2mv²

1/2x 11.8kg x 10.4²

= 11.8x108.16/2

= 1276.288/2

= 638.144 J

This is the translational energy

B.

Rotational energy

1/2Iw²

I = mv²/2

w = v/r

When we substitute

1/2[mv²/2][v/r]²

= 1/4mv²

= 1/4 x 11.8 x 10.4²

= 1276.288/4

= 319.072

This is the rotational energy

C

Total kinetic energy.

We add the values of A and B above to get this.

638.144 J + 319.072 J

= 957.216J