Figure shows a smasll sphereical bal of mass m rolling down the loop track. Thebasll is released on the linear portion at a verticla height H from the lowest point. The circular part show has a radius R.
a. find the kinetic energy of the ball when it is at a point where the radius makes an angle `theta` with the horizontal.
Find the radial and the tangential accelerations of the cente when the ball is at A. c. find the bnormal force and the frictionasl force acting on the ball if H=60 cm, R=10cm `theta=0` and m=70 fg.

Answer

A, B, C, D

Explanation

a. Total kinetic energy
`y=1/2mv^2+1/2Iomega^2`
therefore, according to the question
`mgH=1/2mv^2+1/2Iomega^2+mgR(1+sintheta)`
`rarr mgH-mgR(1+sintheta)`
`=1/2mv^2+1/2Iomega^2`
`rarr 1/2mv^2+1/2Iomega^2=mg(H-R-Rsintheta)`
b. To find the acceleratiohn components,
`rarr 1/2 mv^2+1/2Iomega^2=mg(H-R-Rsintheta)`
`rarr k7/10 mv^2=mg(H-R-Rsintheta)`
`rarr v^2=10/7g(H-R-Rsintheta)`
`rarr v^2=10/7g(H-R-Rsintheta)`.....i
`rarr v^2/R=10/7 (g[H-R)-Rsintheta])/R`
`rarr v^2/R=(10g[(H-R)-Rsintheta)])/R`
`-2v (dv)/(dt)=-(10/7)g R costheta(dtheta)/(dt)`
`rarr omegaR(dv)/(dt)=-(5/7)gRcostheta (dtheta)/(dt)`
`rarr (dv)/(dt)=-(5/7)gcostheta`
`rarr ` tangential acceleration
c. Normal force at `thea=0`
`(mv^2)/R=(70/100)xx(10/7)xx10{((0.6-0.1))/0.1}`
`=5N`
Friction force:
`=f=mg-ma`
`-f=m(g-a)`
`m(10-5/7x10)`
`=0.07(10-5/7xx10)`
`=1/100(70-50)=0.2N`