A man is holding an 820kg dumbell at arms length a distance

A man is holding an 8.20-kg dumbell at arm\'s length, a distance of 0.550 m from his shoulder. What is the torque on the shoulder joint if the arm is held at 40degree below the horizontal? A) 12.6 Nm B) 33.8Nm C) 2.20Nm D) 21.6Nm E) 4.40Nm 22) A meter stick balances at the 50.0-cm mark. If a mass of 39.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm mark. What is the mass of the meter stick? A) 99.1 g B) 73.4 g C) 178 g D) 89.7 g E) 127 g 20) Grumpy cat is stuck riding a merry-go-round, which has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s^2. If he is laying 4.65 m from the center of the merry-go-round, what is the magnitude of Grumpy cat\'s acceleration? A) 2.96 m/s^2 B) 3.23 m/s^2 C) 7.27 m/s^2 D) 8.05 m/s^2 E) 7.30 m/s^2 21) Two woman, waiting for their children to exit school, are chatting while sitting on a merry-go-round. Woman A is at a greater distance from the axis of rotation than woman B. Which of the women has the larger angular speed if the merry-go-round starts to rotate. A) Woman A B) Woman B C) They have the same zero angular speed. D) They have the same non-zero angular speed. E) There is not enough information given to answer the question. 19) Consider a small uniform rod with length 0.157 m long. The rod is released from a vertical position with one end resting on a table. What is the angular speed of the rod when it makes a 30.0degree angle with the vertical (the end on the table does not slip)? A) 3.56 rad/s B) 7.23 rad/s C) 6.32 rad/s D) 9.91 rad/s E) 5.01 rad/s 20) Grumpy cat is stuck riding a merry-go-round, which has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s^2. If he is laying 4.65 m from the center of the merry-go-round, what is the magnitude of Grumpy cat\'s acceleration? A) 2.96 m/s^2 B) 3.23 m/s^2 C) 7.27 m/s^2 D) 8.05 m/s^2 E) 7.30 m/s^2 18) A block is resting on the outer edge of a merry-go-round at a distance of 4.65 m from the center axis. The block has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 rad/s^2. There are two tpes of acceleration, centripetal and angular. What angle does the acceleration (i.e. vector total of the two) of the block make with the tangential direction? A) 64.5degree B)84.1degree C) 90.0degree D) 27.5degree E) 42.5degree

Solution

23) torque = r x F = r F sin@

where @ is angle between F and r , @ - 90 - 40 = 50 deg

     = 0.550 x 8.20 x 9.8 x sin50 = 33.86 N m

Ans(B)

24) balancing moment about balance point ( x = 0.613 m )

[ (0.90 - 0.613) x 39 g ] - [ (0.613 - 0.5) x m g ) ] = 0

0.287 x 39 = 0.113 m

m = 99.1 g

Ans (A)

20 )

tangential acc. a_t = alpha*r = 0.745 x 4.65 = 3.46 m/s^2

radial acc. a_c = w^2 r   = 1.25^2 x 4.65 = 7.26 m/s^2

a = sqrt[ a_t^2 + a_c^2 ] = 8.05 m/s^2

Ans(D)

21) angular speed will be same for every point. Ans(D)