A block of mass m moves on a 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.

Explanation

(a) Normal contact force provides necessary centripetal force
`N=(mv^(2))/(R )`
(b) `f=muN=(mumv^(2))/(R )`
(c ) `f=-ma_(t)`
`(mumv^(2))/(R )=-ma_(t)impliesa_(t)=-(muv^(2))/(R )`
(d) `a_(t)=-(muv^(2))/(R )`
`v(dv)/(ds)=-(muv^(2))/(R )`
`int_(v_(0))^(v)(dv)/(v)=-(mu)/(R ) int_(0)^(2xx2piR) ds`
`|log_(e)v|_(v0)^(v)=-(mu)/(R )xx4piR=-4pimu`
`log_(e)(v)/(v_(0))=4pimuimpliesv=v_(0)e^(-4pimu)`
[Distance traveled along circle=Displacement in the context of tangential direction]