Question

A bowling ball encounters a 0.760-m vertical rise on

theway back to the ball rack, as the drawing illustrates.
Ignorefrictional losses and assume that the mass of the ball
isdistributed uniformly. The translational speed of the ballis 3.50
m/s at the bottom of the rise. Find the translationalspeed at the
top.

Answer 1

Answer:1.26 m/s

Step-by-step explanation:

Given

translation speed of ball =3.5 m/s

Moment of inertia of ball about com
`I=(2)/(5)mr^2`

Initial Energy

`E_i=(1)/(2)mu^2+(1)/(2)I\omega _i^2(\omega =(u)/(r))`

Final Energy

`E_f=(1)/(2)mv^2+(1)/(2)I\omega _f^2+mgh`

Equating energy as no energy loss take place

`E_i=E_f`

`(1)/(2)mu^2+(1)/(2)I\omega _i^2=(1)/(2)mv^2+(1)/(2)I\omega _f^2+mgh`

`(1)/(2)mu^2+(1)/(2)* (2)/(5)mr^2* \left ( (u)/(r)\right )^2=(1)/(2)mv^2+(1)/(2)* (2)/(5)mr^2* \left ( (v)/(r)\right )^2+mgh`

m term get cancel

`\left ( (u^2)/(2)\right )+\left ( (2u^2)/(10)\right )=\left ( (v^2)/(2)\right )+\left ( (2v^2)/(10)\right )+gh`

`(7)/(10)u^2=(7)/(10)v^2+gh`

`v^2=3.5^2-(10)/(7)* 9.81* 0.76`

`v=√(1.6)=1.26 m/s`