A horizontal 805-N - merry-go-round is a solid disk or radius 1.52 m and is star

Question

A horizontal 805-N - merry-go-round is a solid disk or radius 1.52 m and is starred from rest by a constant horizontal force of 50.7 N applied tangentially to the edge of the disk. Find the kinetic energy of the disk J A cylinder of mass 6.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of 9.0 m/s. (a) Determine the translational kinetic energy of its center of mass. J (b) Determine the rotational kinetic energy about its center of mass. You do not know the radius of this cylinder, so work through the equations with a symbol R and see if you need it at the end. What is the relationship between the speed 9.0 and the angular speed? (c) Determine its total energy. J

Explanation / Answer

formula for rotational kinetic energy is

K = 1/2 * I w^2

M = Mass o weight /g = 805 N/9.8m/s2 = 82.14 Kg

= F*R

= (50 .7 N)*1.52 m

(50 .3 N)*1.52 m = ( 82.14Kg)*(1.52 m)2/2 *

=0.805 rad/s2

w = wo+ alpha t

= 0 + 0.805 rad/s^2 ( t)

K = 1/2 * I w^2

= 1/2 ( (  ( 82.14Kg)*(1.52 m)2/2) (  0.805 rad/s^2 ( t))^2

substitue t value you will get final answer

As per guidelines I worked first problem, please post remaining questions in the next post

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