A solid sphere with a diameter of 0.17 m is released from

Step 1 of 3

Part a

We are required to calculate the horizontal distance d by which the ball moves. To calculate the horizontal distance, we need to calculate the linear speed and time considering the linear motion.

Let m be the mass of the sphere.

Using the conservation of mechanical energy principle (Equation 1),

\(mgh=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}\)

where \(v\) is the speed achieved by the ball, \(I\) is the moment of inertia and \(\omega\) is the angular speed.

Given:

\(\text { Diameter of the sphere }=0.17 \mathrm{~m}\)

\(\text { Radius of the sphere } r=0.085 \mathrm{~m}\)

The moment of inertia of  a sphere is

\(I=\frac{2}{5} m r^{2}\)

Height \(h=0.61 \mathrm{~m}\) 

Substitute in equation (1),

\(\begin{array}{l} 9.81 \times 0.61 m=\frac{1}{2} m v^{2}+\frac{1}{2} \times \frac{2}{5} m r^{2} \times \frac{v^{2}}{r^{2}} \\ 5.98=\frac{1}{2} v^{2}+\frac{1}{5} v^{2} \\ v=2.92 \mathrm{~m} / \mathrm{s} \end{array}\)

For a vertical height \(h_{1}=\frac{1}{2} g t^{2}\)

\(\begin{array}{l} t=\sqrt{\frac{2 h}{g}} \\ t=\sqrt{\frac{2 \times 1.22}{9.81}} \\ t=0.5 \mathrm{~s} \end{array}\)

s (the height now is 1.22 m)

Therefore, the horizontal distance \(d=2.92 \times 0.5 \mathrm{~m}\)

\(d \approx 1.5 \mathrm{~m}\)

The required horizontal distance is 1.5 m.