The decrease in potential energy equals the increase in kinetic energy, most of which is rotational KE. First compute the moment of inertia of the yo-yo, I.
If it falls a distance L (which is 1.1 m in this case) then
M g L = (1/2) I w^2 + (1/2) M V^2
The angular rotation rate w can be replaced by V/r, where r is the radius of the cylindrical hub. Then solve for V.
V^2[1 + I/(Mr^2)] = 2*g*L
Take it from there.
A yo-yo is made of two solid cylindrical disks, each of mass 0.053 kg and diameter 0.073 m, joined by a (concentric) thin solid cylindrical hub of mass 0.0052 kg and diameter 0.012 m. Use conservation of energy to calculate the linear speed of the yo-yo when it reaches the end of its 1.1 m long string, if it is released from rest.
2 answers
I still don't get it.Moment of inertia of the yo-yo is 1/2 mr^2: 1/2(.053kg)(.0365^2) + 1/2(.0052kg)(.006^2)
which equals to 3.54*10^-5
so i plugged that into = square root of (2*9.8*1.1)/ (1 + (3.54*10^-5/(mr^2))
and i got something like 4.6 which is not a correct answer. Can you help me clarify this question please?
which equals to 3.54*10^-5
so i plugged that into = square root of (2*9.8*1.1)/ (1 + (3.54*10^-5/(mr^2))
and i got something like 4.6 which is not a correct answer. Can you help me clarify this question please?