# Torsional Couple Problems

A Circular bar 1m long and of 8mm diameter is rigidly clamped at one end in a vertical position. A couple of magnitude of 2.5Nm is applied at the other end. As a result, a mirror fixed at this end deflects a spot of light by 15cm on a scale 1m away. Calculate the rigidity modulus of the bar?
Torque,ζ=(∏n[r][/4]∅)/2l where r is the radius of the rod,l is length,n is the rigidity modulus,∅is the angle of twist.

Applying given values in equation
=>n=4.1*10 raised to 10 N/m square

But the answer provided by this question is 8.3*10 raised to 10 N/m square.

Similarly another problem using same equation

Two cylinders P and Q of radii in the ratio 1:2 are soldered co axially. The free end of P is clamped and the free end of B is twisted by an angle of 17 degree. Determine the twist at the junction,if the cylinders are of the same length and made of the same material?

The answer I got is 17*16,i.e twist=272 degree.

But here also answer provided is wrong from my point of view.It is given only 16 degree.

If I have made any mistake,kindly correct me with explanation.

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CWatters
Homework Helper
Gold Member
Just a guess but did you use the diameter instead of the radius? I've not checked if that's the case. Perhaps show your working.

Just a guess but did you use the diameter instead of the radius? I've not checked if that's the case. Perhaps show your working.

2.5Nm = (∏*n*(4*10 raised to -3)raised to 4*15*10 raised to -2)/2
i.e n = (2.5*2)/(∏*(4*10 raised to -3)raised to 4*15*10 raised to -2)
=>n=4.14*10 raised to 10 N/m square.

This is the answer for 1st question.

For the second.
Let the radius of P be r.Hence from the given ration,the radius of Q will be 2r.Given that length,rigidity modulus,length of both cylinders are EQUAL.Also for this case Torque,ζ is same for 2 cylinders.
For the first cylinder Q,
ζ/∅(q)=(∏n(2r)raised to 4)/2l
ζ/∅(q)=(16∏n(r)raised to 4)/2l ->equation(1) ; {∅(q) is the angle of twist of the cylinder Q}

For the second cylinder P,
ζ/∅(p)=(∏n(r)raised to 4)/2l ->equation(2) ; {∅(p) is the angle of twist of the cylinder P}

From equations (1) & (2),we have

ζ/∅(q)=16* ζ/∅(p)
But ∅(q) = 17 degree,it is given.And cancelling ζ on both sides
1/17=16/∅(p)

i.e ∅(p)=17*16=272 degree.

If there is anything to clear let me know.And if there is any mistake in my calculations or answers provided by the questions are wrong,then kindly inform me.

CWatters
Homework Helper
Gold Member
Perhaps it's me but I'm having a hard time understanding where this equation came from and what is raised to what power..

Torque,ζ=(∏n[r][/4]∅)/2l
How did you get that and can you use the "go advanced" button. Lets you add superscripts.

Perhaps it's me but I'm having a hard time understanding where this equation came from and what is raised to what power..

How did you get that and can you use the "go advanced" button. Lets you add superscripts.
Actually I have no idea how to raise terms to some power here.But I will repeat that
ζ=(∏*n*r raised to 4*∅)/2l
where ζ is the torque acting on the body,∏ is pie or 3.14,n is the Rigidity modulus of the material of the cylinder,r is the radius of it,∅ is the angle of twist,and l is the length of the cylinder.
Actually this is the equation of Torsional Couple. I don't know whether this problem should be posted on "Advanced Section" of this forum.By the way I am undergraduate Physics student.