A small ball of mass 0.75 kg is attached to one end of a 1.25m-long massless rod

Question

A small ball of mass 0.75 kg is attached to one end of a 1.25m-long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 30 degrees from the vertical, what is the magnitude of the torque about the pivot?

I used T=I,

I= 1/2mR2,

I=1/2(.75)(1.25^2)

I=.586

BUT I can't find alpha because none of the rotational speeds were given. Any ideas?

I was also going to use T=rFsin, but there's no force....unless we resolve 1.25m into 1.25sin30 and 1.25cos30 to find the resultant? ...........? Please help :(

Explanation / Answer

Mass of the ball m = 0.75 kg , lengthof the rod = 1.25 m          torque = F l sin                       = mg l sin ( 30 )                       = 0.75 kg * 9.81 m/s2 * 1.25 m * (0.5 )                      = 4.59 N m Mass of the ball m = 0.75 kg , lengthof the rod = 1.25 m          torque = F l sin                       = mg l sin ( 30 )                       = 0.75 kg * 9.81 m/s2 * 1.25 m * (0.5 )                      = 4.59 N m

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