Question

A puck of mass m = 0.065 kg is moving in a circle on a horizontal frictionless surface. It is held in its path by a massless string of length L = 0.65 m. The puck makes one revolution every t = 0.55 s. what is the magnitude of the tension in the string while the pack is revolving?if the string breaks how fast will the pack move away?

Answer 1

The centripetal acceleration of an object in circular motion with tangential velocity v if the radius of the trajectory is r, is:

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

On the other hand, if the period of one revolution is T, then the distance traveled is 2πr (which is the circumference of the trajectory) and the tangential velocity is:

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

In this case, the radius of the trajectory is the length of the string, 0.65m, and the period is T=0.55s. Substitute these data into the formula to find v:

$v=\frac{2\pi(0.65m)}{0.55s}=7.4256\frac{m}{s}$

Plug in v=7.4256m/s and r=0.65m to find the centripetal acceleration of the puck:

$a_c=\frac{(7.4256\frac{m}{s})^2}{0.65m}=84.83\frac{m}{s^2}$

According to Newton's Second Law of Motion, there must be a net force acting on the object responsible for the centripetal acceleration. That force F is the tension of the string. Then:

$F=m\cdot a_c$

Substitute the value for a and m=0.065kg to find the tension of the string:

$\begin{gathered} F=0.065\operatorname{kg}\times84.83\frac{m}{s^2} \\ =5.51N \end{gathered}$

If the string breaks, the puck will move as fast as its tangential velocity, which is 7.4 m/s.

Therefore, the answers are:

Tension in the string: 5.51N

Speed of the puck if the string breaks: 7.43 m/s