Question

A 10-kg cylinder rolls without slipping on a rough surface. at an instant when its center of gravity has a speed of 10 m/s, determine (a) the translational kinetic energy of its center of gravity, (b) the rotational kinetic energy about its center of gravity, and (c) its total kinetic energy.

Answer 1

Part a)

`K_t = 500 J`

Part b)

`K_r = 250 J`

Part c)

`K = 750 J`

Step-by-step explanation:

Part a)

As we know that cylinder is rolling without slipping

So we have

`v = R\omega`

now we have

`v = 10 m/s`

`m = 10 kg`

now we know that translational kinetic energy is given as

`K_t = (1)/(2)mv^2`

`K_t = (1)/(2)(10)(10^2)`

`K_t = 500 J`

Part b)

Now for rotational kinetic energy we know that

`K_r = (1)/(2)I \omega^2`

`K_r = (1)/(2)((1)/(2)mR^2)((v)/(R))^2`

`K_r = (1)/(4)mv^2`

`K_r = (1)/(4)(10)(10^2)`

`K_r = 250 J`

Part c)

Total kinetic energy is given as

`K = K_r + K_t`

`K = 500 + 250`

`K = 750 J`