Friction on rotating object

[diegocas]

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 $$v_0$$ and after a turn it has velocity $$\frac{1}{2}v_0$$.

What is the coefficient of kinetic friction?

Homework Equations

Theorem of Work-Energy:

$$W_{total} = K_f - K_i$$

The Attempt at a Solution

The difference in kinetic energy is easily computed:

$$K_f - K_i = \frac{1}{2} m (\frac{1}{2}v_0)^2 - \frac{1}{2}m v_0^2 = -\frac{3}{8}m v_0^2$$

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:

$$W_{total} = W_{friction} = - F_{friction} \Delta x = - (\mu_k m g) (2 \pi r)$$

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?

 

[anathema]

Hello, thank you for your post. It seems like you are on the right track with your calculations. To calculate the coefficient of kinetic friction, we can use the equation for work done by friction:

W_{friction} = - \mu_k N \Delta x

Where N is the normal force and \Delta x is the displacement along the surface. In this case, the normal force is equal to the weight of the object, mg, and the displacement \Delta x is equal to the circumference of the circle, 2 \pi r.

As for the direction of the frictional force, it is always opposite the direction of motion. In this case, the motion is circular, so the frictional force will always act tangentially to the circle, towards the center. This is necessary for the object to maintain a uniform circular motion. If the frictional force had a radial component, it would cause the object to accelerate or decelerate, and it would not be able to maintain a constant velocity.

I hope this helps clarify any confusion. Keep up the good work!