Superhard Potential energy/circular motion problem

[brendan3eb]

Homework Statement

A ball is tied to a cord and set in rotation in a vertical circle. Prove that the tension in the cord at the lowest point exceeds the tension in the cord at the highest point by six times the weight of the ball.

Homework Equations

U1+K1=U2+K2
F=mv^2/r
a=-V^2/r

The Attempt at a Solution

From U1+K1=U2+K2 I get:
(1/2)mv^2=2mgr+(1/2)mv2^2

I also took a look at the forces to get:
T1=mg-mv^2/r
and
T2=-mv^2/r-mg

I have tried substituting all sorts of ways, but have been unable to prove what is asked for. I asked my teacher for help, but he said that I was on the right track and I just had to keep plugging stuff in.

 

[TMM]

It seems to me like it would only differ by twice the weight of the ball.

By the way, check your values for the tension, remembering that tension is a scalar.Oh I see; nvm I was being dumb.

 

[PhanthomJay]

Homework Statement

A ball is tied to a cord and set in rotation in a vertical circle. Prove that the tension in the cord at the lowest point exceeds the tension in the cord at the highest point by six times the weight of the ball.

Homework Equations

U1+K1=U2+K2
F=mv^2/r
a=-V^2/r

The Attempt at a Solution

From U1+K1=U2+K2 I get:
(1/2)mv^2=2mgr+(1/2)mv2^2

I also took a look at the forces to get:
T1=mg-mv^2/r
and
T2=-mv^2/r-mg

I have tried substituting all sorts of ways, but have been unable to prove what is asked for. I asked my teacher for help, but he said that I was on the right track and I just had to keep plugging stuff in.

Your plus and minus signs are all messed up. At the top of the circle, the centripetal acceleration is down, therefore, the net centripetal force must be down, and T and mg act in the same direction (down). At the bottom of the circle, the centripetal acceleration is upward, the net force must be up, and since T acts up and mg acts down, T must be greater than mg. Continue to use Newton 2 and conservation of energy, and keep on plugging once you've corrected your signage. The answer you are trying to prove is correct.

 

[brendan3eb]

I corrected the signs to get

T1=m(g+v1^2/r)
and
T2=m(V2^2/R-g)

I set T1 over T2 and I plugged 4gr + v2^2 in for V1^2 to get:

(5g+v2^2/r)/(v2^2/r-g)

I am still plugging in different values right now, but I am posting this because I think I may have the wrong expression as nothing seems to be working out.

 

[PhanthomJay]

I corrected the signs to get

T1=m(g+v1^2/r)
and
T2=m(V2^2/R-g)

I set T1 over T2 and I plugged 4gr + v2^2 in for V1^2 to get:

(5g+v2^2/r)/(v2^2/r-g)

I am still plugging in different values right now, but I am posting this because I think I may have the wrong expression as nothing seems to be working out.

You now have the right signage and the right equations, and now it's just algebra. You are trying to show that T1 - T2 = 6mg. So don't set T1 over T2; rather subtract T2 from T1 and use your substitution for v1^2 and you will get
T1 -T2 = 6mg.