A ball of mass 0.25kg attached to the string of length 1.96m is moving in a horizontal circle. The string will break if the tension in the string is more than 25N. What is the maximum angular velocity with which the ball can be moved?
A
7.14rad/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
3.92rad/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.57rad/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
14.28rad/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A7.14rad/s The tension in the string provides centripetal acceleration to the ball. The relation between the tension T and the angular velocity ω is given by mω2r=T,
where m is mass of the ball and r is length of the string. The maximum angular velocity occurs at the maximum tension (breaking tension) in the string i.e., ωmax=√Tmaxmr=√25(0.25)(1.96)=7.14rad/s