Posted by Joe on Wednesday, July 11, 2012 at 7:09pm.
A ball (mass m = 250 g) on the end of an ideal string is moving in a circular motion as a conical pendulum. The length L of the string is 1.84 m and the angle with the vertical is 37 degrees.
a) What is the magnitude of the torque (N m) exerted on the ball about the support point?
b) What is the magnitude of the angular momentum (kg m^2/2) of the ball about the support point?
Correct Answers: a) 2.71 b) 1.32
For a I assumed 0 because there wasn't any said force. I do not know how to solve problem.
For b I used L = m*v*r where L = momentum
m = .250 kg
v = (r * g * tan 37 )^(0.5) = (L*sin 37*9.8*tan 37)^(0.5) = 2.8596
r = L * sin 37 = 1.84 * sin 37= 1.10733
Therefore L = .791 This was supposedly incorrect

Physics  bobpursley, Wednesday, July 11, 2012 at 7:14pm
a. torque= LXforce, force is mg downward.
torque= 1.84*.250*9.8*sin(18037)

Physics  Joe, Wednesday, July 11, 2012 at 7:50pm
Any idea how I was hung up on 2nd one?

Physics  Elena, Thursday, July 12, 2012 at 5:50am
(a) There two forces acting on the ball: gravity and tension. The line of tension psses through the support point => its torque about this point is zero.
The totque of gravity about support point is
τ=mg•Lsinα=0.25•9.8•1.84•sin37º= 2.71 N•m
(b)
x and yprojections of acting forces are
x: m•v²/R= T•sinα, (1)
y: m•g=T•cosα. (2)
Divide (1) by (2)
m•v²/R•m•g = T•sinα/ T•cosα
v=sqrt(R•g•tanα) =
=sqrt(L•sinα• g•tanα)=
sqrt(1.84•9.8•0.6•0.75) =2.86 m/s.
Angular momentum =m•v•L=
0.25•2.86•1.84=1.32 kg•m²/s.
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