An unusual yoyo is constructed from a hollow sphere of mass M and radius R, and

Question

An unusual yoyo is constructed from a hollow sphere of mass M and radius R, and is allowed to fall while unrolling a massless string attached to the ceiling.

a. Draw the free-body diagram for the yoyo during the fall. Include any coordinate system(s) you will use in the following part.

b. Write Newtons Second Law in either or both the translational form (F=ma) and the rotational form (?=I?) as appropriate to allow you to solve for the motion of the sphere.

c. What is the magnitude of the acceleration of the center of mass in terms of M, R, and the acceleration of gravity, g, or a subset of these quantities?

d. What is the magnitude of the strings tension in terms of M, R, and the acceleration of gravity, g, or a subset of these quantities?

An unusual yoyo is constructed from a hollow sphere of mass M and radius R, and is allowed to fall while unrolling a massless string attached to the ceiling. a. Draw the free-body diagram for the yoyo during the fall. Include any coordinate system(s) you will use in the following part. b. Write Newton's Second Law in either or both the translational form (F=ma) and the rotational form (?=I?) as appropriate to allow you to solve for the motion of the sphere. c. What is the magnitude of the acceleration of the center of mass in terms of M, R, and the acceleration of gravity, g, or a subset of these quantities? d. What is the magnitude of the string's tension in terms of M, R, and the acceleration of gravity, g, or a subset of these quantities?

Explanation / Answer

the free body diagram has two forces and one net acceleration
a force of tension acting upwards
weight of yoyo acting downwards
so Using Newton's formula f = ma
mg - T = ma

The torque acting on the body is from the tension in the rope at radius r. (not same as R, radius if disk)
the weight acts through centre of mass and produces no torque on the body)
Using Euler's formula tau (torque) = I (inertia) * alpha (angular acc)
mR^2/2 * a / r = T*r
where r is 0.5m
We have two unknowns T and a, and two equations, so we can solve for them
we can write
T = m *a*(R/r)^2
ans so
mg - m*a*(R/r)^2 = m*a
a = g / (1 + (R/r)^2 )
So we know the acceleration which is constant.

we now need standard straight line motion equations of bodies with constant acceleration
Using the face that it started from rest and falls two meters
v^2 = 2 *a *1m (from v^2 - u^2 = 2*a*S)
get v
and then for time
t = v / a (from v = u + a*t)
where a is obtained in terms of g, R, r

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