A wheel rotating about a fixed axis with a constant acceleration of 3.0 rad/s^2

Question

A wheel rotating about a fixed axis with a constant acceleration of 3.0 rad/s^2 turns through 3.7 revolutions. If the wheel starts at rest what is the angular velocity at the end of this time interval? a. 9.5 rad/s b. 14 rad/s c. 12 rad/s d. 3.2 rad/s e. 8.8 rads Five identical cylinders viewed from the end are each acted on by forces of equal magnitude. If each can only spin about their center symmetry axis, in which configuration will the force will exert the biggest torque? A ball of mass 3m at x = 0 is connected to a ball of mass m at x = L by a massless rod. Consider the three rotation axes A, B and C as shown, all parallel to the y axis. Consider three rotating axes A, B, and C, all parallel to the y axis, and I_A, I_B and I_C are the moment of inertia for each axis, respectively. The rotation takes place is in the x-z plane. Compare the three moment of inertia. a. I_A > I_B > I_C b. I_C = I_A > I_B c. I_C > I_B > I_A d. I_B > I_C = I_A e. I_C > I_B = I_A

Explanation / Answer

12.

Angular acceleration,=3.0 rad/s2

Angular displacement,=3.7 rev=(3.7)(2)(3.14)=23.236 rad.

Initial angular velocity =0

Final angular velocity is given by

2=2 =2(3)(23.236)

=11.81=12 rad/s

13.

by definition of torque.

torque=force(perpendicular distance)

so torque in figure a is maximum.

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