- #1

- 16

- 0

## Homework Statement

An object of mass m lies on a rough surface and is tied by a rope to a fixed point. It rotates in a circle of radius r around the fixed point. Initially the object has velocity [tex] v_0 [/tex] and after a turn it has velocity [tex] \frac{1}{2}v_0 [/tex].

What is the coefficient of kinetic friction?

## Homework Equations

Theorem of Work-Energy:

[tex] W_{total} = K_f - K_i [/tex]

## The Attempt at a Solution

The difference in kinetic energy is easily computed:

[tex] K_f - K_i = \frac{1}{2} m (\frac{1}{2}v_0)^2 - \frac{1}{2}m v_0^2 = -\frac{3}{8}m v_0^2 [/tex]

However, how am I supposed to calculate the work of the frictional force?

It is easy under the assumption that the frictional force is always opposite the velocity. In this case:

[tex] W_{total} = W_{friction} = - F_{friction} \Delta x = - (\mu_k m g) (2 \pi r) [/tex]

However, why can't the frictional force have a radial component?

For example, if I wanted to keep the object in a uniform circular motion, the frictional force would have to be radial.

Any insights?