Weight causing a wheel to rotate

[JamesGold]

Homework Statement

Homework Equations

I_solidcylinder = .5mr^2
I_thinring = mr^2
torque = I*Ω (where Ω = angular acceleration)
change in theta = .5(Ω)(time^2)

The Attempt at a Solution

Attempted to solve for angular acceleration:

(2.5)(9.8)(0.1) = (3*0.25^2 + 0.5*4*0.1^2)Ω

Ω = 11.8 rad/s/s

which I then plugged into the rotational kinematics formula:

Δ∅ = 0.5*11.8*0.5^2

which gave me a wrong answer. Where'd I go wrong?

 

[grzz]

The torque causing angular acceleration is provided by the tension in the string.
But is this tension equal to the weight of the hanging mass?
Note that the hanging mass is accelerating downwards.

 

[JamesGold]

But is this tension equal to the weight of the hanging mass?

I don't see why it wouldn't be. Can you explain why it isn't?

 

[JHamm]

What if you had two blocks connected by a string and dropped them over a balcony, would the string have a tension equal to the weight of the lower block?

 

[grzz]

Is the hanging mass accelerating downwards?

 

[JamesGold]

What if you had two blocks connected by a string and dropped them over a balcony, would the string have a tension equal to the weight of the lower block?

I'm not sure. Wouldn't it depend on the relative weights of the two blocks?

Is the hanging mass accelerating downwards?

Yes.

 

[grzz]

So if the hanging mass is accelerating which is the greater force, the tension upwards or the weight of the mass downwards?

 

[JamesGold]

The weight of the mass downwards.

 

[grzz]

Correct.

 

[JamesGold]

So mg - T = ma?

 

[grzz]

Hence the tension is not equal to the weight.

The torque is provided by the tension and this tension is still unknown.

But one can obtain another equation so that one can find the tension.

 

[grzz]

So mg - T = ma?

exactly.

 

[JamesGold]

Then how do you find the acceleration of the block?

 

[grzz]

The linear acceleration is related to the angular acceleration.

Note that

for distance there is, x = rθ
hence for velocity there is, v = rω
and for acceleration there is, a = rα

 

[grzz]

I used α instead of Ω for angular acc

 

[JamesGold]

Right... but without knowing the tension of the string, I don't see how we can calculate the angular acceleration of the wheel or the acceleration of the block.

 

[grzz]

We have two equations:

1...torque eqn for rotational acc

2...f = ma for linear acc

get linear acc 'a' in terms of the ang acc

and then eliminate tension from these two equations.

 

[JamesGold]

Okay, so I have now:

String tension * moment arm = I of system * ang acc

[ (2.50)(9.80) - (2.50)(0.1)α ] * 0.1 = (0.02 + 0.1875) * α

Is this correct?

 

[grzz]

Yes.

 

[JamesGold]

Got the right answer. I can't thank you enough, selfless sir.

 

[grzz]

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