Show every step for the following problems. 1. Consider a motorcycle of mass 150
ID: 1514934 • Letter: S
Question
Show every step for the following problems.
1. Consider a motorcycle of mass 150. kg, one wheel of which has a mass of 10. kg and a radius of 30. cm. What is the ratio of the rotational kinetic energy of the wheels to the total translational kinetic energy of the bike? Assume the wheels are uniform disks.
Answer: 0.13 : 1
2. Two children, each of mass 20.0 kg, ride on the perimeter of a small merry-go-round that is rotating at the rate of 1.0 revolution every 4.0 s. The merry-go-round is a disk of mass 30. kg and radius 3.0 m. The children now both move halfway in toward the center to positions 1.5 m from the axis of rotation. Calculate the kinetic energy before and after, and the final rate of rotation. Explain any changes in energy
Answer:
Ko = 0.61 kJ
Kf = 1.3 kJ
f = 0.55 Hz
Children do work moving inward.
Explanation / Answer
(1)
Total Transaltional Kinetic Energy = 1/2 * M*v^2
Total Rotational Kinetic Energy = 1/2 * I * ^2
Where,
I = 1/2 * m*r^2
= v/r
Rotational Kinetic Energy / Transaltional Kinetic Energy = (2*1/2 * 1/2 * m*r^2 * v^2/r^2)/(1/2 * M*v^2)
Rotational Kinetic Energy / Transaltional Kinetic Energy = (1/2 * 10) / (1/2 * 150)
Rotational Kinetic Energy / Transaltional Kinetic Energy = 0.067 / 1
(2)
= 1/4 rev/s
= 1.57 rad/s
K.E before = 2* 1/2*Ic*^2 + 1/2 * Id * ^2
K.E before = 2* 1/2 * 20.0 * 3.0^2 * 1.57^2 + 1/2 * 1/2* 30 * 3.0^2 * 1.57^2
K.E before = 610.1 J
K.E before = 0.61 KJ
Finding Final Angular velcoity, Using Momentum Conservation,
2*Ic*i + Id *i = 2*Ic*f + Id *f
2* 20.0 * 3.0^2 * 1.57 + 1/2* 30 * 3.0^2 * 1.57 = 2 * 20.0 * 1.5^2 * f + 1/2* 30 * 3.0^2 * f
f = 3.45 rad/s
K.E after = 2* 1/2*Ic*f^2 + 1/2 * Id * f^2
K.E before = 2* 1/2 * 20.0 * 1.5^2 * 3.45^2 + 1/2 * 1/2* 30 * 3.0^2 * 3.45^2
K.E before = 1339 J
K.E before = 1.3 KJ
Final rate of rotaion, 2**f = 3.45
f = 3.45/(2*)
f = 0.55 Hz