# Rotational Motion About a Fixed Axis

1. Nov 27, 2007

### wchvball13

1. The problem statement, all variables and given/known data
Two thin rectangular sheets (0.23 m 0.35 m) are identical. In the first sheet the axis of rotation lies along the 0.23 m side, and in the second it lies along the 0.35 m side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in 6.5 s. How long does it take for the second sheet, starting from rest, to reach the same angular velocity?

2. Relevant equations
$$\tau=(mr^{2})\alpha$$
$$E\tau=I\alpha$$
$$I=(1/3)ML^{2}$$

3. The attempt at a solution
I don't know how to start if you don't have a mass...or the angular acceleration

Last edited: Nov 27, 2007
2. Nov 27, 2007

### rl.bhat

w = wo + alpha*t
Torque = I*alpha
Alpha = Torque/I
wo = 0 Torque is same, final angular velocity is same.
Therefore (Torque/I1)t1 = (Torque/I2)t2
I1 = (M* 0.35^2)/3 and I2 = ( M*0.23^2)/3

Last edited: Nov 28, 2007
3. Nov 27, 2007

### Staff: Mentor

The point is that the sheets are identical and the same torque is applied, but the moment of inertias will be different because of the different orientation, which means a different value for L in the expression for moment of inertia.

4. Nov 28, 2007

### wchvball13

I understand both of your replies, but I still don't understand how I can solve it if I don't have M or alpha
I1 = (M* 0.35^2)/3 and I2 = ( M*0.23^2)/3

5. Nov 28, 2007

### rl.bhat

w = wo + alpha*t.....(1)
Torque = I*alpha....(2)
Alpha = Torque/I....(3)
wo = 0 Torque is same, final angular velocity is same.
Therefore (Torque/I1)t1 = (Torque/I2)t2......(4)
I1 = (M* 0.35^2)/3 and I2 = ( M*0.23^2)/3
From eq. 2 you can find alpha. Put this value in eq.1. Put wo = 0. and equate w for I1 and I2. In the final expression M gets cancelled out.

6. Nov 29, 2007

### wchvball13

apparently I suck at physics because I can't even understand how you can find Alpha from eq. 2. Once I figure out that I can find the rest but for right now all I have is I1=.0408 and I2=.0176....and that's if I ignore the masses since they're equal...

7. Nov 29, 2007

### rl.bhat

From eq.4 you get t1/I1 = t2/I2, because torque is same. t1 = 6.5 s. Find t2.
The text passage contains a physics problem related to rotational motion, specifically dealing with the time it takes for two identical rectangular sheets to reach the same angular velocity under the same torque but with different axes of rotation.
single_question_with_answer

the passage includes a detailed problem statement, relevant equations, and a discussion where the user and helpers work through the solution. the question is clearly stated, and the answer (or the method to find the answer) is provided in the discussion.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- MULTIPLE_QUESTIONS
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system
# Rotational Motion About a Fixed Axis

1. Nov 27, 2007

### wchvball13

1. The problem statement, all variables and given/known data
Two thin rectangular sheets (0.23 m 0.35 m) are identical. In the first sheet the axis of rotation lies along the 0.23 m side, and in the second it lies along the 0.35 m side. The same torque is applied to each sheet. The first sheet, starting from rest, reaches its final angular velocity in 6.5 s. How long does it take for the second sheet, starting from rest, to reach the same angular velocity?

2. Relevant equations
$$\tau=(mr^{2})\alpha$$
$$E\tau=I\alpha$$
$$I=(1/3)ML^{2}$$

3. The attempt at a solution
I don't know how to start if you don't have a mass...or the angular acceleration

Last edited: Nov 27, 2007
2. Nov 27, 2007

### rl.bhat

w = wo + alpha*t
Torque = I*alpha
Alpha = Torque/I
wo = 0 Torque is same, final angular velocity is same.
Therefore (Torque/I1)t1 = (Torque/I2)t2
I1 = (M* 0.35^2)/3 and I2 = ( M*0.23^2)/3

Last edited: Nov 28, 2007
3. Nov 27, 2007

### Staff: Mentor

The point is that the sheets are identical and the same torque is applied, but the moment of inertias will be different because of the different orientation, which means a different value for L in the expression for moment of inertia.

4. Nov 28, 2007

### wchvball13

I understand both of your replies, but I still don't understand how I can solve it if I don't have M or alpha
I1 = (M* 0.35^2)/3 and I2 = ( M*0.23^2)/3

5. Nov 28, 2007

### rl.bhat

w = wo + alpha*t.....(1)
Torque = I*alpha....(2)
Alpha = Torque/I....(3)
wo = 0 Torque is same, final angular velocity is same.
Therefore (Torque/I1)t1 = (Torque/I2)t2......(4)
I1 = (M* 0.35^2)/3 and I2 = ( M*0.23^2)/3
From eq. 2 you can find alpha. Put this value in eq.1. Put wo = 0. and equate w for I1 and I2. In the final expression M gets cancelled out.

6. Nov 29, 2007

### wchvball13

apparently I suck at physics because I can't even understand how you can find Alpha from eq. 2. Once I figure out that I can find the rest but for right now all I have is I1=.0408 and I2=.0176....and that's if I ignore the masses since they're equal...

7. Nov 29, 2007

### rl.bhat

From eq.4 you get t1/I1 = t2/I2, because torque is same. t1 = 6.5 s. Find t2.
user
The text passage contains a physics problem related to rotational motion, specifically dealing with the time it takes for two identical rectangular sheets to reach the same angular velocity under the same torque but with different axes of rotation.

TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage includes a detailed problem statement, relevant equations, and a discussion where the user and helpers work through the solution. The question is clearly stated, and the answer (or the method to find the answer) is provided in the discussion.