Thank you for registering.

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

Please check your email for login details.
MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A srone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is in the lowest position , and has a speed u. The magnitude of the change in its velocity as it reaches a position where the sting is horizontal is : a) root (u^2 - 2gL) b) root (2gL) c) root (u^2 - gL) d) root (2(u^2 - gL)) please help by answering this Q and the method to do this Question

A srone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is in the lowest position , and has a speed u. The magnitude of the change in its velocity as it reaches a position where the sting is horizontal is :


a) root (u^2 - 2gL)


b) root (2gL)


c) root (u^2 - gL)


d) root (2(u^2 - gL))


 


please help by answering this Q and the method to do this Question

Grade:12

7 Answers

vikas askiitian expert
509 Points
10 years ago

applying energy conservation

 1/2mu^2=mgl+1/2mv^2

  we get v=root(u^2-2gl)

  in vector form we can express as v vector=root(u^2-2gl)j and u vector =ui

  difference is v-u =root(u^2+v^2)=root(2(u^2-gl))

Sudheesh Singanamalla
114 Points
10 years ago

Can some1 explain this more clearly !

Raiwat Bapat
13 Points
3 years ago
student-name Abhiram M V asked in PhysicsA stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude of the change in the velocity as it reaches a position where the string is horizontal is (A) u2−2gL−−−−−−−√ (B) 2gL−−−√ (C) u2−gL−−−−−−√ (D) 2(u2−gL)−−−−−−−−−√ 7 Follow 0student-name A K Daya Sir answered this2879 helpful votes in Physics, Class XI-ScienceLet the direction of revolution be counter-clockwiseInitial velocity = u i^ (at bottom)i^ ( i-cap suggests that the direction of initial velocity is towards positive x-axis)--When the string is horizontal, the body has covered a quarter of revolution, its velocity is v j^j^ (j-cap suggests that the direction of its velocity is towards positive y-axis when the string is horizontal)-Change in velocity = vj^ - ui^magnitude of change in velocity = √ v2 + u2-Using conservation of energyKinetic Energy of body when it is at bottom = 1/2 mu2When it goes to the quarter position, it does some work against gravity. So a part of its kinetic energy is spent in doing work equal to mgh and remaining kinetic energy is 1/2 mv21/2 mu2 = 1/2 mv2 + mghhere h = l1/2 mu2 = 1/2 mv2 + mglv2 - u2 = -2glv2 - u2 + 2u2 = 2u2 - 2glv2 + u2 = 2 (u2 - gl)Therefore, change in magnitude of velocity = √ v2 + u2 = √2 (u2 - gl)
Mohit laxkar
13 Points
2 years ago
 From enegy conservation we kniw that change in potential enegy is equal to that of change in kinetic energy i.e. energy can neither be created bor be destroyed
Δu=ΔK
mgl=1/2mu2 -1/2mv2
gl= u2/2 - v2/2
v=√u2-2gl
vishu
34 Points
2 years ago
The portion of Velocity change –
change in velocity = v jcap – u icap
and the magnitude of change in velocity =>
whole uder-root ( Vsquare + U square + 2 * V * U * Cos90 )
=> whole under-root (Vsquare+Usquare +0 )                           (as cos90 = 0)
=> whole under-root (Vsquare+Usquare) 
 
So, we have to find the value of (Vsquare+Usquare) ,i.e., 
whole under-root   [2*{U square – gl }]    (ans.)
Mukesh Shukla
15 Points
2 years ago
A stone tied to a string of lenght L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed miu. the magnitude of the change in its velocity as it reaches a position where the string is horizontal is:
Rishi Sharma
askIITians Faculty 646 Points
11 months ago
Dear Student,
Please find below the solution to your problem.

applying energy conservation
1/2mu^2=mgl+1/2mv^2
we get v=root(u^2-2gl)
in vector form we can express as v vector
=root(u^2-2gl)j and
u vector =ui
difference is v-u
= root(u^2+v^2)
= root(2(u^2-gl))

Thanks and Regards

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free