Answer to Question #128036 in Physics for Catherin Romero

Question

cylinder of m= 0.53 kg and a radius 0.32 m is rotating around an axis through its center. the cylinder’s moment of inertial is given by MR^2/2. The cylinder initially at rest. A force of 0.70n is then applied tangentially at the edge of the cylinder. How many radians will the cylinder rotate through in the first 2.5 seconds that the force is applied?

Answer 1

The Newton's second law for rotation is the following:

$$\varepsilon = T/I$$

where $\varepsilon$ is the angular acceleration, $I = \dfrac{mr^2}{2}$ is the moment of inertia and $T$ is the torque of the force $F = 0.7N$. To find the torque we should multiply the force by the cilinder radius:

$$T = Fr$$

Thus, obtain the following angular acceleration:

$$\varepsilon = \dfrac{Fr}{mr^2/2} = \dfrac{2F}{mr} = \dfrac{2\cdot 0.7}{0.53\cdot 0.32} \approx 8.25\space rad/s^2$$

On the other hand, the kinematic law for the rotation with the constant acceleration (starting from the rest) is the following:

$$\theta = \dfrac{\varepsilon t^2}{2}$$

Here $\theta$ is the amount of radians the cylinder will rotate through in the time $t = 2.5s$.

Thus, obtain:

$$\theta = \dfrac{8.25\cdot 2.5^2}{2} \approx 25.8 \space rad$$

Answer. 25.8 rad.