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.

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  • Ash
    Lv 7
    6 months ago
    Favorite 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

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