Home
Class 11
PHYSICS
In the above question, ifu=sqrt(gR) then...

In the above question, `ifu=sqrt(gR)` then
(a) after rotating an angle `theta` , velocity of the bob becomes zero. Find the value of `theta` .
(b) If mass of the bob is `'m'` then what is the tension in the string when velocity becomes zero?

Text Solution

Verified by Experts

The correct Answer is:
B
(a) Velocity becomes zero at height `h_(2)` discussed in article `10.5` where, `h_(2)=(u^(2))/(2g)=(gR)/(2g)=(R)/(2)`
Now, `h_(2)=R(1-costheta)`
:. `(R)/(2)=R(1-costheta) impliestheta=60^(@)`
(b) `T=mgcostheta=(mg)/(2)attheta=60^(@)`
Promotional Banner

Topper's Solved these Questions

  • CIRCULAR MOTION

    DC PANDEY|Exercise Assertion And Reason|28 Videos

  • CIRCULAR MOTION

    DC PANDEY|Exercise Level 1 Objective|14 Videos

  • CIRCULAR MOTION

    DC PANDEY|Exercise Exercise 10.2|7 Videos

  • CENTRE OF MASS, LINEAR MOMENTUM AND COLLISION

    DC PANDEY|Exercise Level 2 Subjective|21 Videos

  • COMMUNICATION SYSTEM

    DC PANDEY|Exercise Only One Option is Correct|27 Videos

Similar Questions

Explore conceptually related problems

Suppose the amplitude of a simple pendulum having a bob of mass m is theta_0 . Find the tension in the string when the bob is at its extreme position.

Let theta denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the bob is m, the tension in the string is mg cos theta

The velocity of a body of mass m revolving in a vertical circle of radius R at the lowest point 2sqrt2gR . The minimum tension in the string will be

A pendulum bob of mass m and length L is released from angle theta with the vertical. Find (a) the speed of the bob at the bottom of the swing and (b) tension in the string at that time

A block of mass M si suspended by two strings angles theta_(1) and theta_(2) with the horizontal. Find the tensions in the strings.

In question number -1,ifu=sqrt(7gR) then ltbr (a) What is the velocity at topmost point? (b) What is tension at topmost point? (c) What is tension at bottommost point?

The bob of a simple pendulum is displaced from its equilibrium position 'O' to a position 'Q' which is at a height 'h' above 'O' and the bob is then released. Assuming the mass of the bob to be 'm' and time period of oscillation to be 2.0s , the tension in the string when the bob passes through 'O' is

(a) Two charged particles , each charge q , are joined by an insultating string of length L and system is kept on smooth surface. Find tension in the string. (b) A particle having charge Q_(0) is placed directly below and at separation d from the bob of simple pendulum at rest. The mass of the bob is m . What charge should be given to bob so that string becomes loose ? ( c) A particle A having charge q and mass m is placed at the bottom of a smooth inclined plane of inclination theta . Where should a block B , having same charge and mass , be placed on the incline so that it may remain in equilibrium ?