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October 24, 2014

October 24, 2014

Posted by **Joe** on Wednesday, July 11, 2012 at 7:09pm.

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 7:14pma. torque= LXforce, force is mg downward.

torque= 1.84*.250*9.8*sin(180-37)

- Physics -
**Joe**, Wednesday, July 11, 2012 at 7:50pmAny 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 y-projections 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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