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Constant Angular Acceleration

  • Thread starter blue5t1053
  • Start date
23
1
Question:
A flywheel has a constant angular acceleration of 2 rad/sec^2. During the 19 sec time period from t1 to t2 the wheel rotates through an angle of 15 radians. What was the magnitude of the angular velocity of the wheel at time t1?

Hint: let t1=0 sec, and t2=t

Equations:
[tex]\vartheta - \vartheta_{0} = \omega_{0} t + \alpha t ^{2}[/tex]

My Work:
[tex](15 radians) - (0 radians) = \omega_{0} (19 sec) + (2 \frac{rad}{sec^{2}})(19 sec )^{2}[/tex]

[tex](15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2} = \omega_{0} (19 sec)[/tex]

[tex]\frac{(15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2}}{(19 sec)} = \omega_{0}[/tex]

[tex]\omega_{0} = (-18.2)\frac{rad}{sec} = (18.2)\frac{rad}{sec} \ for \ magnitude; \ at \ t1[/tex]

Did I do everything right?
 
Last edited:

Answers and Replies

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,474
20
Question:
A flywheel has a constant angular acceleration of 2 rad/sec^2. During the 19 sec time period from t1 to t2 the wheel rotates through an angle of 15 radians. What was the magnitude of the angular velocity of the wheel at time t1?

Hint: let t1=0 sec, and t2=t

Equations:
[tex]\vartheta - \vartheta_{0} = \omega_{0} t + \alpha t ^{2}[/tex]

My Work:
[tex](15 radians) - (0 radians) = \omega_{0} (19 sec) + (2 \frac{rad}{sec^{2}})(19 sec )^{2}[/tex]

[tex](15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2} = \omega_{0} (19 sec)[/tex]

[tex]\frac{(15 radians) - (2 \frac{rad}{sec^{2}})(19 sec )^{2}}{(19 sec)} = \omega_{0}[/tex]
You're good up to here.

[tex]\omega_{0} = (-18.2)\frac{rad}{sec} = (18.2)\frac{rad}{sec} \ for \ magnitude; \ at \ t1[/tex]
You've miscalculated. You should get something close to 37 rad/s for the magnitude.
 
102
0
I think your equation is wrong

Should it not be

theta(final) - theta(initial) = time * angular velocity(initial) + (1/2) * angular acceleration * time^2.

similar to the equation in linear motion?
 
Doc Al
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