A particular motor can provide a maximum of 110.0 N · m of torque.Assuming that all of this torque is used to accelerate a solid,uniform flywheel of mass 10.0 kg and radius 3.00 m,how long will it take for the flywheel to accelerate from rest to 8.13 rad/s?
A)3.33 s
B)2.83 s
C)4.03 s
D)4.36 s

Question

Quizplus · Accepted Answer

Explanation:The torque acting on the flywheel is:$$	au = I \alpha$$where I is the moment of inertia of the flywheel and \alpha is the angular acceleration.The moment of inertia of a solid, uniform flywheel is:$$I = \frac{1}{2}MR^2$$where M is the mass of the flywheel and R is its radius.Substituting the given values into these equations, we get:$$	au = \frac{1}{2}(10.0 	ext{ kg})(3.00 	ext{ m})^2(8.13 	ext{ rad/s}^2) = 373.4 	ext{ N} \cdot 	ext{m}$$$$\alpha = \frac{	au}{I} = \frac{373.4 	ext{ N} \cdot 	ext{m}}{\frac{1}{2}(10.0 	ext{ kg})(3.00 	ext{ m})^2} = 8.13 	ext{ rad/s}^2$$The angular acceleration is constant, so we can use the following equation to find the time it takes for the flywheel to accelerate from rest to 8.13 rad/s:$$\omega_f = \omega_i + \alpha t$$where \omega_i is the initial angular velocity (0 rad/s), \omega_f is the final angular velocity (8.13 rad/s), and t is the time.Solving for t, we get:$$t = \frac{\omega_f - \omega_i}{\alpha} = \frac{8.13 	ext{ rad/s} - 0 	ext{ rad/s}}{8.13 	ext{ rad/s}^2} = 1.00 	ext{ s}$$Therefore, it will take 1.00 s for the flywheel to accelerate from rest to 8.13 rad/s.

A Particular Motor Can Provide a Maximum of 110