An green hoop with mass m h = 2.8 kg and radius R h = 0.12 m hangs from a string

Question

An green hoop with mass mh = 2.8 kg and radius Rh = 0.12 m hangs from a string that goes over a blue solid disk pulley with mass md = 2 kg and radius Rd = 0.09 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms = 4.1 kg and radius Rs= 0.24 m. The system is released from rest.

1)What is magnitude of the linear acceleration of the hoop?

m/s2

2)What is magnitude of the linear acceleration of the sphere?

m/s2

3)What is the magnitude of the angular acceleration of the disk pulley?

rad/s2

4)What is the magnitude of the angular acceleration of the sphere?

rad/s2

5)What is the tension in the string between the sphere and disk pulley?

N

6)What is the tension in the string between the hoop and disk pulley?

N

7)The green hoop falls a distance d = 1.52 m. (After being released from rest.)

How much time does the hoop take to fall 1.52 m?

s

8)What is the magnitude of the velocity of the green hoop after it has dropped 1.52 m?

m/s

9)What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.52 m)?

rad/s

I need help with this question plezzzzzz :(

Explanation / Answer

the only force on the system is the weight of the hoop
F_net = 2.8kg*9.81m/s2 = 27.468 N

The inert masses that must be accelerated are the ring, pulley, and the rolling sphere.

The mass equivalent of M the pulley is found by
torque = F*R = I* = I*a/R
F = M*a = I*a/R2 -->
M = I/R2 = 21/2*m*R2/R2 = 1/2*m

The mass equivalent of the rolling sphere is found by:
the sphere rotates around the contact point with the table.
So using the theorem of parallel axes, the moment of inertia of the sphere
is I = 2/5*mR2 for rotation about the center of mass + mR2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR2
M = 7/5*m

the acceleration is then
a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4.1) = 27.468/9.54 = 2.879 m/s2
----------------
1)linear acceleration of hoop= 2.879 m/s2

2) lin. accel. of sphere = 2.879 m/s2

3) angular acceleration disk pulley:
= a/R = 2.879/0.09 = 31.99 rad/s2

4) ang. acceleration sphere = a/R = 2.879/0.24 = 12 rad/s2

5) tension between pulley and sphere = M*a = 7/5*4.1*2.879 = 16.5254 N

6) tension between hoop and pulley = m(hoop) (g - a)
= 2.8(9.81 - 2.879) = 19.4 N

7) s = 1/2*at^2 => t = sqrt(2s/a) = sqrt(2*1.52/2.879)
=> t = 1.027 seconds

8) v = sqrt(2as) = sqrt(2*2.922*1.52) = 2.98 m/s

9) = v/R = 2.98/0.24 = 12.41 rad/s

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