Question

A heavy (2.0 kg) point-like object rests 2.0m from the center of a rough turntable as the turntable rotates. The period of the turntable's rotation is 5.0 seconds. The coefficient of kinetic friction between the object and turntable is 0.50, while the coefficient of static friction is 0.80. What is the magnitude of the force of friction acting on the object?

Answer 1

Answer:

6.32N

Explanation:

According to Newton's second law:

$\sum F=ma$

In this case the only force that acts on the object is the friction force, and the acceleration, is the centripetal acceleration since it is a circular movement, so we have:

$F_f=ma_c(1)$

Centripetal aceleration is given by:

$a_c=\frac{v^2}{r}(2)$

The speed is given by:

$v=\omega r\\\omega=\frac{2\pi}{T}\\v=\frac{2\pi r}{T}$

Replacing $v$ in (2) and $a_c$ in (1):

$F_f=m\frac{v^2}{r}\\F_f=m\frac{(\frac{2\pi r}{T})^2}{r}\\F_f=m\frac{4\pi^2 r}{T^2}\\F_f=2kg(\frac{4\pi^2(2m)}{(5s)^2)}\\F_f=6.32N$