A block os mass m moves on as horizontal circle against the wall of a cylindrical room of radius R. The floor of the room onwhich the block moves is smoth but the friction coefficient between the wall and the block is `mu`. The block is given an initial speed `v_0`. As a function of the speed v write a. the normal force by the wall on the block. b. thefrictional force by the wall and c. the tangential acceleration of the block. d. Integrate the tangential acceleration `((dv)/(dt)=v(dv)/(ds))` to obtain the speed of the block after one revoluton.

Answer

A, B, C, D

Explanation

A block of mass m mioves on a horizontla circle against the wall of a cylindrical room of radius R.
Friction coefficient between wall annd the block is m.
a. Normal reaction by the wall on the block is `=(mv^2)/r`
b. `:. FrictioN/Al force by wall
`=(mumv^2)/r`
c. `(mumv^2)/R=ma`
rarr a=-(muv^2)/R` (Deceleration)
d. Now ((dv)/(dt)=v((dv)/(ds))=-(muv^2)/R`
`rarr ds=-r/mu (dv)/v`
`rarr s=-R/mu In v+c`
At s=0, `v=v_0`
Therefore, `c=R/muIn v_0`
So, `s=-R/mu In v/v_0`
`rarr v/v_0=e^((mus)/R)`
`rarr For one rotation
`s=2piR`
`So, v=v_0e^(-2pimu)`