Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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. .
initially total energy (TE) = mu2/2 + PE = mu2/2 (PE of ground is taken 0)
=mu2/2
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)
thnx 4 helping
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !