Question

A ball of mass $ 0.25\, kg $ attached to the end of a string of length $ 1.96\, m $ is moving in a horizontal circle. The string will break if the tension is more than $ 25 \,N $ . What is the maximum speed with which ball can be moved?

• A. $ 14 \,m/s $
• B. $ 3 \,m/s $
• C. $ 3.92 \,m/s $
• D. $ 5 \,m/s $

Answer 1

The Correct Option is A

The correct option is(A):  14m/s.
Tension of string \(T=\frac{m v^{2}}{r}\) \(25 =\frac{0.25 \times v^{2}}{1.96}\) \(v =14\, m / s\)

Answer 2

Explanation: 

According to the question, the string will break under the condition where the force exerted, reaching beyond 25N, which is the upward force that is binding the string and the ball together. This upward force is the tension in the string. The outward force that is acting on the ball is the centripetal force. If the centripetal force is greater than the upward tension, only then the string will break. This means we have to equate the centripetal force and tension of the string, to attain the maximum frequency that the ball can acquire.

When an object moves in circular motion, centripetal force acts outwards, therefore:

\(F=\frac{mv^2}{r}\)

where, 

F = centripetal force,

m = mass of the body revolving around the circular motion,

r = radius of the circle formed due to circular motion,

v = velocity of the body

Given: mass of ball (m) = 0.25 kg,

radius (r) = 1.96m,

F = 25 N

Substituting the values in the above equation, we get:

\(=>25=\frac{0.25v^2}{1.96}\)

v²=196

=>v-14

Therefore, the maximum speed with which the ball can be moved is 14m/s.

 

Answer 3

To find the maximum speed with which the ball can be moved without breaking the string, we need to consider the forces acting on the ball in circular motion.
Given:
- Mass of the ball, \( m = 0.25 \, \text{kg} \)
- Length of the string, \( r = 1.96 \, \text{m} \)
- Maximum tension in the string, \( T = 25 \, \text{N} \)The tension in the string provides the centripetal force required to keep the ball moving in a horizontal circle. The centripetal force \( F_c \) is given by:
\[ F_c = \frac{mv^2}{r} \]Where \( v \) is the speed of the ball.Since the tension \( T \) is the maximum centripetal force the string can withstand, we set \( F_c = T \):
\[ T = \frac{mv^2}{r} \]Solving for \( v \):
\[ v^2 = \frac{Tr}{m} \]
\[ v = \sqrt{\frac{Tr}{m}} \]Plugging in the given values:
\[ v = \sqrt{\frac{25 \times 1.96}{0.25}} \]
\[ v = \sqrt{\frac{49}{0.25}} \]
\[ v = \sqrt{196} \]
\[ v = 14 \, \text{m/s} \]Therefore, the maximum speed with which the ball can be moved without breaking the string is \( \boxed{14 \, \text{m/s}} \).