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Chapter 8: Q. 54 (page 201)

FIGURE P8.54 shows a small block of mass m sliding around the inside of an L-shaped track of radius r. The bottom of the track is frictionless; the coefficient of kinetic friction between the block and the wall of the track is µk,The block's speed is vo at to =0 Find an expression for the block's speed at a later time t.

Short Answer

Expert verified

The expression for the velocity at time t is vt=(vor)(voμkt+r)

Step by step solution

01

Given Information

mass= m

radius= r

The bottom of the track is frictionless

coefficient of kinetic friction between the block and the wall of the track is µk,

speed is vo at to =0

02

Explanation

Centripetal force is given by

Fc=mv2r
where m = mass, v = velocity and r = radius.
Frictional force is µkN

Where µk,= coefficient of friction and N is Normal force

From the Newton's second law, we know

mdvdt=-μk(mv2r)dvdt=-μkv2r

Now integrate and find vt

∫v0vt-dvv2=∫0tμkdtr1vt-1v0=μktr1vt=μktr+1v0vt=vorvoμkt+r

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