A block is mass m moves on as horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth 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. the frictional 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)`