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pezzang

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Hi, thank you so much for making this kind of post for students like me. I have tried the quesotins but i want to make sure that i was right. if worng, please correct me.

THANK YOU SO MUCH. Also i had attached pictures of the question to give you enough information. PLEASE REFER TO THE PICTURE ATTACHED.

Q) The two uniform disks shown above have equal mass, and each can rotate on frictionless bearings about a fixed axis through its center. The smaller disk has a radius R and moment of inertia I about its axis.

(a) determine the moment of inertia of the larger disk about tis axis in terms of I.

-> I used the equation Inertia = mass*radius to solve this question,

Let's say that smaller disk has a subscript "A" and larger; "B".

Then, I(B) = m(2R) = mR(2).

since mR is equal to I, we have I(B) = I*2 = 2I

so the answer is 2*I. Am I right?

The two disks are then linked as shown below by a light chain that cannot slip. They are at rest when, at time t = 0, a student applies a torque to the smaller disk, and it rotates counterclockwise with constant angular acceleration alpha. Assume that the mass of the chain and the tension in teh lower part of the chain are negligible.

In terms of I, R, alpha and t, determine each of the following.

(b) The angular acceleration of the larger disk (alpha(B))

-> torque(A) = I*alpha

torque(B) = 2I*alpha(B)

torque(A) = torque(B) <- IS IT POSSIBLE THAT I ASSUME THIS?

I*alpha = 2I*alpha(B)

alpha(B) = alpha / 2. <- AM I RIGHT?

(c) The tension in the upper part of the chain

->

T(Tension) = (I*alpha) / R.

T(A) = (I*alpha) / R

T(B) = (2I)(alpha(B)) / (2R) = (2I*alpha/2) / R = (I*(alpha) / (2R)

TO FIND TENSION, DO I SUBTRACT THE SMALLER ONE FROM LARGER ONE??

IF I DO, I GET:

T(A) = T(B) = (I*alpha) / R - (I*alpha) / (2R) = (I*alpha) / (2R)

Therefore, the tension in teh upper part of the chain equals to (I*(alpha) / (2R).

(d) The torque that the student applied to the smaller disk

->

torque(A) = I*alpha <- I found this equation in my physics book. Is right? I kind of doubt it though. Please help me.

(e) The rotational kinetic energy of the smaller disk as a function of time

->

RKE = (1/2)*I*w^2 (omega)

to find an equation of w(omega) in terms of t and alpha, we take antiderivative of the equation alpha = (dw)/(dt), then we get:

alpha*t = w

IF we substitute alpha*t for w(omega) in the equation RKE = (1/2)*I*w^2 (omega), we get

RKE = (1/2)I*(alpha*t)^2.

So that is the answer.

So please correct me and thank you so much. Have a wonderful day, people!

THANK YOU SO MUCH. Also i had attached pictures of the question to give you enough information. PLEASE REFER TO THE PICTURE ATTACHED.

Q) The two uniform disks shown above have equal mass, and each can rotate on frictionless bearings about a fixed axis through its center. The smaller disk has a radius R and moment of inertia I about its axis.

(a) determine the moment of inertia of the larger disk about tis axis in terms of I.

-> I used the equation Inertia = mass*radius to solve this question,

Let's say that smaller disk has a subscript "A" and larger; "B".

Then, I(B) = m(2R) = mR(2).

since mR is equal to I, we have I(B) = I*2 = 2I

so the answer is 2*I. Am I right?

The two disks are then linked as shown below by a light chain that cannot slip. They are at rest when, at time t = 0, a student applies a torque to the smaller disk, and it rotates counterclockwise with constant angular acceleration alpha. Assume that the mass of the chain and the tension in teh lower part of the chain are negligible.

In terms of I, R, alpha and t, determine each of the following.

(b) The angular acceleration of the larger disk (alpha(B))

-> torque(A) = I*alpha

torque(B) = 2I*alpha(B)

torque(A) = torque(B) <- IS IT POSSIBLE THAT I ASSUME THIS?

I*alpha = 2I*alpha(B)

alpha(B) = alpha / 2. <- AM I RIGHT?

(c) The tension in the upper part of the chain

->

T(Tension) = (I*alpha) / R.

T(A) = (I*alpha) / R

T(B) = (2I)(alpha(B)) / (2R) = (2I*alpha/2) / R = (I*(alpha) / (2R)

TO FIND TENSION, DO I SUBTRACT THE SMALLER ONE FROM LARGER ONE??

IF I DO, I GET:

T(A) = T(B) = (I*alpha) / R - (I*alpha) / (2R) = (I*alpha) / (2R)

Therefore, the tension in teh upper part of the chain equals to (I*(alpha) / (2R).

(d) The torque that the student applied to the smaller disk

->

torque(A) = I*alpha <- I found this equation in my physics book. Is right? I kind of doubt it though. Please help me.

(e) The rotational kinetic energy of the smaller disk as a function of time

->

RKE = (1/2)*I*w^2 (omega)

to find an equation of w(omega) in terms of t and alpha, we take antiderivative of the equation alpha = (dw)/(dt), then we get:

alpha*t = w

IF we substitute alpha*t for w(omega) in the equation RKE = (1/2)*I*w^2 (omega), we get

RKE = (1/2)I*(alpha*t)^2.

So that is the answer.

So please correct me and thank you so much. Have a wonderful day, people!

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