An green hoop with mass m_h = 2.4 kg and radius R_h = 0.15 m hangs from a string

Question

An green hoop with mass m_h = 2.4 kg and radius R_h = 0.15 m hangs from a string that goes over a blue solid disk pulley with mass m_d = 2.4 kg and R_d = 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 m_s = 4 kg and radius R_s = 0.18 m. The system is released from rest. What is magnitude of the linear acceleration of the hoop? What is magnitude of the linear acceleration of the sphere? What is the magnitude of the angular acceleration of the disk pulley? What is the magnitude of the angular acceleration of the sphere? What is the tension in the string between the sphere and disk pulley? What is the tension in the string between the hoop and disk pulley? The green hoop falls a distance d = 1.6 m. (After being released from rest.) How much time does the hoop take to fall 1.6 m? What is the magnitude of the velocity of the green hoop after it ha

Explanation / Answer

First, the hoop. There's no indication that it rotates, so "F=ma". The hoop has gravity (mhg) pulling down, and the tension in the vertical string (call it "Tv") pulling up.
Fnet = ma
mhg Tv = mha

Next, the pulley. It only rotates, has no translational motion. So skip the "F=ma" and just do the " = I".
p = Ipp
First, we know that the only torques on the pulley are due to the vertical string's tension (Tv) pulling one way, and the horizontal string's tension (call it "Th") pulling the other way. So:
p = TvRd ThRd = (Tv Th)Rd
So:
(Tv Th)Rd = Ipp
Also, we know that the moment of inertia for a uniform disk is ½mR², so:
(Tv Th)Rd = ½(md)(Rd)²()
Also, since the string is not slipping, the edge of the pulley is moving at the same speed as the string, the hoop, and the sphere. So this means its angular acceleration (p) is related to the string's linear acceleration (a) by the relationship "p = a/Rd". So:
(Tv Th)Rd = ½(md)(Rd)²(a/Rd)

Tv Th = ½mda

Next, the orange sphere.
There's a certain amount of tension (Th) pulling it one way (toward the pulley), but there's also a force of friction (call it "Ff") pulling it the opposite way. So:
Fs = msa
Th Ff = msa
Consider the axis through the center, then the "Th" force doesn't contribute to the torque (because its line of action goes through the axis of rotation). But the "Ff" force does produce a torque, in the amount of FfRs. So:
s = FfRs = Iss
For a uniform sphere, I = (2/5)mR², so:
FfRs = (2/5)(msRs²s)
Also, s = a / Rs, so:
FfRs = (2/5)(msRs²a/Rs)
Simplify:
Ff = (2/5)(msa)
Now substitute for "Ff" in the force equation above, to get:
Th (2/5)(msa)  = msa
Simplify:
Th = (7/5)(msa)

Now we have these equations:
mhg Tv = mha
Tv Th = ½(mda)
Th = (7/5)(msa)

Solving these equations,

a = mhg/((1/2)md + (7/5)ms + mh) = 2.56 m/s2

(1) So the magnitude of the linear acceleration of the hoop is 2.56 m/s2.

(2) Since linear acceleration of the sphere is same as that of hoop so its magnitude is 2.56 m/s2.

(3) Angular acceleration of the disk pulley, p = a/Rd =2.56/0.09 = 28.41 rad/s2.

(4) Angular acceleration of the sphere, s = a/Rs=2.56/0.18 = 14.22 rad/s2.

(5) Tension in the string between sphere and the disk pulley, Th =(7/5)msa =7*4*2.56/5 = 14.34 N.

(6) Tension in the string between disk pulley and hoop, Tv =(g-a)mh =(9.8-2.56)*2.4 =17.38 N.

(7) Initial velocity of hoop, uh = 0

Say final velocity of hoop is vh

then, vh = at, (t=time)

and (vh)2 = 2aS, (S=distance)

From above equations, t=1.12 s

So The hoopm takes 1.12 s to fall 1.6 m.

(8) vh = at =2.56*1.12 = 2.86 m/s

Magnitude of the velocity of the hoop after it has dropped 1.6 m is 2.86 m/s.

(9) magnitude of the final angular speed of orange sphere = linear speed of sphere/Rs = linear speed of hoop after 1.6 m fall/Rs = 2.86/0.18 = 15.9 rad/s

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