Question

A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.5 m on a frictionless horizontal surface. If the cord will break when the tension in it exceeds 63 N, what is the maximum speed the ball can have?

Answer 1

Concepts and reason

The concepts required to solve the question is Newton’s 2^nd law.

Initially apply the Newton’s 2^nd law of motion to find the force acting on the ball. Then, as the ball is rotated, centripetal acceleration will act on it. Replace the acceleration in the equation of force by the centripetal acceleration to find the maximum speed the ball can have.

Fundamentals

The Newton’s 2nd law of motion is given by,

$F = ma$

Here, F is the force acting on the body, m is mass of the body, v is the velocity, t is the time for which the velocity is changing, and a is the acceleration.

The force acting on the ball is given by,

$F = ma$

Here, F is the force acting on the ball, m is the mass of the ball, and a is the acceleration.

The centripetal acceleration acting on the ball is given by,

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

Thus, the force acting on the body is given by,

$F = m\left( {\frac{{{v^2}}}{r}} \right)$

Rearrange the above equation to find the value of v.

$\begin{array}{c}\\{v^2} = \frac{{rF}}{m}\\\\v = \sqrt {\frac{{rF}}{m}} \\\end{array}$

Substitute 0.40 kg for m, 1.5 m for r, and 63 N for F in the equation of v.

$\begin{array}{c}\\v = \sqrt {\frac{{\left( {1.5\;{\rm{m}}} \right)\left( {63\;{\rm{N}}} \right)}}{{\left( {0.40\;{\rm{kg}}} \right)}}} \\\\ = \sqrt {236.25\;{{\left( {{\rm{m/s}}} \right)}^2}} \\\\ = \pm 15.4\;{\rm{m/s}}\\\end{array}$

The maximum speed the ball can have is 15.4 m/s.

Ans:

The maximum speed the ball can have is 15.4 m/s.