# Physics help?

Refer to diagram 4.

A hollow ball (mass 7.97 kg, radius 62.4 cm) is rolling along a horizontal surface at initial speed 5.63 m/s when it comes to a descending incline and then a lower horizontal surface, 76.5 cm below. Assume there are no frictional losses, and the ball never slips. Find v, the linear speed in m/s, of the ball when it rolls along the lower surface.

### 1 Answer

- AshLv 76 months agoFavorite Answer
Total Energy at lower surface = Total energy at the top

linear KE₂ + rotational KE₂ = PE₁ + linear KE₁ + rotational KE₁

½mv₂² + ½Iω₂² = mgh + ½mv₁² + ½Iω₁²

½mv₂² + ½I(v₂/r)² = mgh + ½mv₁² + ½I(v₁/r)²

½v₂²(m + I/r²) = mgh + ½v₁²(m + I/r²)

Multiplying both sides by 2/(m + I/r²), we get

v₂² = 2mgh/(m + I/r²) + v₁²

For hollow sphere I = (2/3)mr², then we get

v₂² = 2mgh/(m + (2mr²/3r²)) + v₁²

v₂² = 2mgh/(m + (2m/3)) + v₁²

v₂² = 2mgh/(5m/3) + v₁²

v₂² = (6gh/5) + v₁²

v₂ = √[(6gh/5) + v₁²]

v₂ = √[(6*9.81*0.624/5) + (5.63)²]

v₂ = 6.25 m/s

Linear speed at lower surface is 6.25 m/s