A bowling ball encounters a 0.760-m vertical rise on the way back to the ball ra

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 7.32 m/s at the bottom of the rise. Find the translational speed at the top.

Number Units No units 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 7.32 m/s at the bottom of the rise. Find the translational speed at the top. No units°skgmm/sNJrad/srad/s^2N·mkg·m^2 Units Number

Explanation / Answer

Answer:-

Given;- Vertical Rise (h) = 0.760 m

Initial velocity (vo)= 7.32 m/s

The total Energy of the ball is at height h and velocity v is

E = Ekin + Epot ----- (1)

kinetic energy is:-

Ekin = Rotational K.E + Translational K.E

=1/2 mv2 + 1/2 I*omega2

The moment of intertial of the ball is:
I = 2/5 mR2

Omega = v/R

Using I and Omega in Ekin we get

Ekin = 1/2 mv2 + 1/2* 2/5 mR2 * (v/R)2

= 7/10 m v2

Potential energy is

Epot = mgh

Using Ekin and Epot in equation 1

E = Ekin + Epot

E = 7/10 mv2 + mgh

The kinetic energy cannot be negative. The sum of potential and kinetic energies is constant. So the decrease in total KE as it goes to the top of the ramp equals the increase in potential energy, MgH.
(7/10)[vo2 - v12) = gH

=> (7/10) [(7.32 m/s)2 - v12) = 9.81 m/s2X 0.76 m

=> 0.7 [53.58 - v12 ] = 7.45

=> 53.58 - v12 = 7.45 / 0.7

=> v12 = 53.58 - 10.65

=> v12 = 42.929

=> v1= 6.55 m/s

So, the translational speed at the top is 6.55 m/s

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