Question

A solid ball (I = 2/5 MR^2) has a mass of 87.0 g and a diameter of 17.0 cm. It is initially at reston a long inclined plane that makes an angle of 25° with respect to the horizontal. If theball rolls down the slope without slipping, calculate the speed of the ball after it has rolled3.00 m down the slope.

Answer 1

Given,

The moment of inertia of the solid ball, I=(2/5)MR²

The mass of the ball, m=87.0 g=0.087 kg

The diameter of the ball, d=17.0 cm=0.17 m

The initial velocity of the ball, u=0 m/s

The angle of inclination of the slope, θ=25°

The distance covered by the ball, s=3.00 m

The acceleration of the ball is given by,

`a=g\sin \theta`

Where g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} a=9.8*\sin 25^(\circ) \\ =4.14\text{ m/s}^2 \end{gathered}`

From the equation of the motion,

`v^2=u^2+2as`

Where v is the speed of the ball after it has rolled 3.00 m

On substituting the known values.

`\begin{gathered} v^2=0+2*4.14*3.00 \\ \Rightarrow v=\sqrt[]{24.84} \\ v=4.98\text{ m/s} \end{gathered}`

Thus the speed of the ball after it rolls 3.00 m down the slope is 4.98 m/s