Question

A 240 N sphere 0.20 m in radius rolls, without slipping 6.0 m downa

Answer 1

Answer:

The angular speed of the sphere at the bottom of the hill is 31.39 rad/s.

Explanation:

It is given that,

Weight of the sphere, W = 240 N

Radius of the sphere, r = 0.2 m

Angle with the horizontal, $\theta=28^{\circ}$

We need to find the angular speed of the sphere at the bottom of the hill if it starts  from rest.

As per the law of conservation of energy, the total energy at the top is equal to the energy at the bottom.

Gravitational energy = translational energy + rotational energy

So,

$mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

I is the moment of inertia of the sphere, $I=\dfrac{2}{5}mr^2$

Also, $v=r\omega$

h is the height of the ramp, $h=l\ sin\theta$

$mgl\ sin\theta=\dfrac{1}{2}m(r\omega)^2+\dfrac{1}{2}I\omega^2$

On solving the above equation we get :

$\omega=\sqrt{\dfrac{10gl\ sin\theta}{7r^2}}$

$\omega=\sqrt{\dfrac{10\times 9.8\times 6\ sin(28)}{7(0.2)^2}}$

$\omega=31.39\ rad/s$

So, the angular speed of the sphere at the bottom of the hill is 31.39 rad/s. Hence, this is the required solution.