Question

A 3.20-kilogram toy car is attached to a 125-centimeter string ofnegligible mass. The other end of the string is attached to a ringthat is slipped over a peg. The tensile strength of the string is144 newions. The car revolves about the peg in a circular path ona frictionless table. Find the maximum angular speed that the carcan travel in revolutions per minute without breaking the string.

Answer 1

Given data:

The tensile strength of the string is,

`F^(\prime)=144\text{ N}`

The mass of the car is,

`m=3.2\text{ kg}`

The length of the string is,

`\begin{gathered} r=125\text{ cm} \\ r=1.25\text{ m} \end{gathered}`

The force acting on the car is,

`F^(\prime)=(mv)/(r)^2`

where a is the linear acceleration and m is the mass of the car.

Substituting the known values,

`\begin{gathered} 144=(3.2* v^2)/(1.25) \\ v^2=56.25 \\ v=7.5ms^(-1) \end{gathered}`

Thus, the angular speed of the car is,

`\begin{gathered} \omega=(v)/(r) \\ \omega=(7.5)/(1.25) \\ \omega=6 \end{gathered}`

Each distance of each revolution is,

`\begin{gathered} d=2\pi r \\ d=2\pi*1.25 \\ d=7.85 \end{gathered}`

Thus, the value of angular speed in terms of revolution is,

`\begin{gathered} \omega=(6)/(7.85) \\ \omega=0.76\text{ revolution per second} \\ \omega=0.76*60\text{ revolutions per minute} \\ \omega=45.6\text{ revolutions per minute} \end{gathered}`

Thus, the maximum angular speed of the toy car is 45.6 revolutions per minute.