A uniform cylinder (I = mR²)of diameter 0.20 m and mass 12 kg rolls without slipping down a 37º inclined plane.The acceleration of the cylinder down the plane is approximately A)2.0 m/s² B)3.9 m/s² C)4.9 m/s² D)5.8 m/s² E)9.8 m/s²

Question

Quizplus · Accepted Answer

We can use the equations for rotational and translational motion to solve this problem. The torque on the cylinder is given by $	au = I\alpha$, where $\alpha$ is the angular acceleration. Since the cylinder is rolling without slipping, we have $a = R\alpha$, where $a$ is the linear acceleration and $R$ is the radius of the cylinder. The torque on the cylinder due to gravity is $	au = MgR\sin	heta$, where $M$ is the mass of the cylinder and $	heta$ is the angle of the incline. Therefore, we have
$$I\alpha = MgR\sin	heta$$
$$a = R\alpha$$
Substituting $I = \frac{1}{2}MR^2$ and solving for $a$, we get
$$a = \frac{2}{3}g\sin	heta$$
Substituting $	heta = 37^\circ$ and $g = 9.8\,	ext{m/s}^2$, we get $a \approx 3.9\,	ext{m/s}^2$. Therefore, the answer is (B).

A Uniform Cylinder (I = MR2)of Diameter 0

A Uniform Cylinder (I = MR²)of Diameter 0