Question

A 1.34 kg ball is connected by means of two massless strings, each of length L 1.70 m, to a vertical, rotating rod. The strings are tied to the rod with separation d = 1.70 m and are taut. The tension in the upper string is 35 N. What are the (a) tension in the lower string, (b) magnitude of the net force on the ball, and (c) speed of the ball? (d) What is the direction of ?

Answer 1

Answer:

The ball and the strings are shown in the attached figure

Balancing the forces in Y direction we have

$35sin(\theta )=1.34\times 9.81+T_{2}sin(\theta )\\\\sin(\theta )=\frac{0.85}{1.7}=0.5\\\\\therefore T_{2}=\frac{35sin(\theta )-13.14N}{sin(\theta )}\\\\T_{2}=8.72N$

The magnitude of net force on ball in

1) Y direction is zero since there is no acceleration of ball along y direction.

2)In x direction is given by $35cos(30)+T_{2}cos(30)=37.88N$

c)

The force in the x direction provides centripetal acceleration for the ball thus we have

$35cos(30)+T_{2}cos(30)=37.88N\\\\37.88=\frac{mv^{2}}{r}\\\\\therefore v=\sqrt{\frac{37.88\times r}{m}}\\\\r=1.70cos(30)=1.47m\\\\\therefore v=\sqrt{\frac{37.88\times 1.47}{1.34}}=6.44m/s$