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blue5t1053
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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?
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?
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