6. A small ball of mass 0.85 kg is attached to one end of a 1.40 m long massless

Question

6. A small ball of mass 0.85 kg is attached to one end of a 1.40 m long massless rod, and the other end of the rod is hung from a pivot. When the resulting pendulum is 40° from the vertical, what is the magnitude of the torque about the pivot? N·m

7. (a) If R = 12 cm, M = 570 g, and m = 60 g (below), find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)
m/s

(b) Repeat (a) with R = 5.0 cm.
m/s

Explanation / Answer

6. I get the Torque = 7.496 Newton meters directed clockwise around the pivot point
The force is 8.33 newtons = m* g
T= FR(sin 37) where R = 1.40 meters, F = 8.33 Newtons, the angle between R and F is 40 degrees
T =(8.33)(1.40)(sin 40)= (7.33)(0.6018)= 7.496 Newton meters

7. Initial Gravitational Potential Energy equals Final Translational Kinetic Energy plus Final Rotational Kinetic Energy:

GPE_i = TKE_f + RKE_f

mgh = ½mv² + ½I²
mgh = ½mv² + ½( ½MR² )²
mgh = ½mv² + ½( ½MR² )( v / R )²
mgh = ½mv² + ¼Mv²
mgh = ( ½m + ¼M )v²

v² = mgh / [ ½m + ¼M ]
v² = ( 0.060 kg )( 9.8 m/s² )( 0.50 m ) / [ ½( 0.060 kg ) + ¼( 0.570 kg ) ]
v² = 0.294 J / [ 0.1725 kg ]
v² = 1.704 m²/s²
v = 1.305 m/s

b. same as A

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