Question

A bowling ball encounters a 0.760-m vertical rise on the way back to the ball rack, as the drawing illustrates. Ignore frictional losses and assume that the mass of the ball is distributed uniformly. The translational speed of the ball is 9.14 m/s at the bottom of the rise. Find the translational speed at the top.

Answer 1

Using Energy conservation at the bottom and top in the given scenario:

KEi + PEi = KEf + PEf

PEi = 0, since initially at the bottom h = 0

PEf = m*g*h, where h = height from ground = 0.760 m

m = mass of bowling ball

KE = total initial kinetic energy = KEtrans + KErot

KE = (1/2)*m*V^2 + (1/2)*I*w^2

I = moment of inertia of ball = 2*m*R^2/5

R = radius of bowling ball

V = translational speed

w = angular velocity = V/R

KE = (1/2)*m*V^2 + (1/2)*(2*m*R^2/5)*(V/R)^2 = (1/2)*m*V^2 + (1/5)*m*V^2

KE = 7*m*V^2/10

Now at the bottom

KEi = 7*m*Vi^2/10, where Vi = 9.14 m/sec

KEf = 7*m*Vf^2/10,

where Vf = final translational speed at the top = ?

So Using above values:

KEi + PEi = KEf + PEf

7*m*Vi^2/10 + 0 = 7*m*Vf^2/10 + m*g*h

divide by m

7*Vi^2/10 + 0 = 7*Vf^2/10 + g*h

Vf^2 = Vi^2 - 10*g*h/7

Using known values:

Vf = sqrt (9.14^2 - 10*9.81*0.760/7)

Vf = 8.54 m/sec = final translational speed at the top