Question

(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord can withstand a maximum tension of 64.0 N, a. What is the maximum speed at which the ball can move before the cord breaks? Assume the string remains horizontal during the motion. (5 marks) b. Suppose the ball moves in a circle of larger radius at the same speed v. Is the cord more likely or less likely to break? ( i ) . A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long . The ball moves in a horizontal circle . If the cord can withstand a maximum tension of 64.0 N , a . What is the maximum speed at which the ball can move before the cord breaks ? Assume the string remains horizontal during the motion . ( 5 marks ) b . Suppose the ball moves in a circle of larger radius at the same speed v . Is the cord more likely or less likely to break ?​

Answer 1

(a) Let $v$ be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

$a_{\rm rad} = \dfrac{v^2}R$

where $R$ is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

$F = (1.500\,\mathrm{kg}) a_{\rm rad}$

so that

$(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N$

Solve for $v$ :

$v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}$

(b) The net force equation in part (a) leads us to the relation

$F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}$

so that $v$ is directly proportional to the square root of $R$. As the radius $R$ increases, the maximum linear speed $v$ will also increase, so the cord is less likely to break if we keep up the same speed.