17. A hoop of mass, m 0.4 kg, rolls without slipping on a horizontal surface so

Question

17. A hoop of mass, m 0.4 kg, rolls without slipping on a horizontal surface so that its center proceeds to the right with a constant linear speed of, v 20 m/s. What is the total kinetic energy of the hoop? (a) 80J (b) 40J (c) 320J (d) 640J (e) 160 18. In the problem above, what is the instantaneous speed at the very bottom of the hoop- where the hoop meets the ramp? (a) zero m/s (b) 20 m/s (c) 40 ms d) 80 mis (e) 160 m's 19. A solid cylinder of radius, R0.25 m, is released from rest& drops a height of, h 4.9 m. It rolls without slipping down the incline as shown. What is the angular speed of the cylinder when it reaches the horizontal surface? (a) 32 rad/s (b) 1024 rad/'s (c) 55.4 rad/s (d) 3073.3 rad's (e) This cannot be determined because the mass is unknown. Hartley

Explanation / Answer

17. mass m = 0.4 kg velocity = 20m/s

total kinetic energy = 0.5 Iw2 + 0.5 mv2

= 0.5 mr2w2 + 0.5 mv2

= 0.5 m r2 v2/r2 + 0.5 mv2 since v = rw

= mv2

= 0.4 (20)2

= 160 J

18. a) zero

19. R = 0.25m h = 4.9m

from law of conservation of energy

mgh = 0.5mv2 + 0.5 Iw2

m gh = 0.5 mv2 + 0.5 (0.5 mr2 w2)

gh = 0.5 r2w2 + 0.25 r2w2

= 0.75 r2w2

w2 = gh/0.75r2

= 9.8 x 4.9 / 0.75 x 0.25 x0.25

= 1024.43

hence w = 32 rad/s

v2 = gh/0.75 = 9.8 x 4.9/0.75 = 36.015

hence v = 6 m/s

angular speed w = v/r = 6 / 0.25

= 24 rad/s

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