badge image

Enroll For Free Now & Improve Your Performance.

User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12


A heavy particle hanging from a fixed point by a light inextensible sting of length 'l' is projected horizontally with speed (gl) 1/2 . Find the speed of the particle and the inclination of the string to the vertical at the instant of the motion when the tension in the string equal to the weight of the particle. .

9 years ago

Answers : (2)

vikas askiitian expert
509 Points

initially total energy (TE) = mu2/2 + PE = mu2/2           (PE of ground is taken 0)



when it sustends @ with verticle then let its velocity is v ...


finally total energy = mv2/2 + PE


PE = mgR(1-cos@) , so total energy finally


TE = mv2/2 + mgR(1-cos@)          .............2


since total energy is conserved so eq1 = eq2

 mv2/2 + mgR(1-cos@) = mu2/2          ................3


now let tention in string at this pos is T then

T - mgcos@ = mv2/R        

T = mg     (given) , so

mv2 = (mg-mgcos@)R       ............4

solving eq 4 & 3

3(1-cos@) = 1

 cos@ = 2/3

@ = cos-1(2/3)



9 years ago
Aiswarya Ram Gupta
35 Points

thnx 4 helping

9 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details