Question:

A solid cylinder of mass 2 kg and radius 0.5 m is initially at rest on a horizontal surface. A constant force of 10 N is applied tangentially to the cylinder's edge, causing it to start rotating. The cylinder accelerates uniformly and reaches an angular velocity of 5 rad/s after 4 seconds.

a) Calculate the moment of inertia of the cylinder. b) Determine the torque applied to the cylinder. c) Find the angular acceleration of the cylinder. d) Calculate the work done by the applied force in 4 seconds. e) Determine the total kinetic energy of the cylinder after 4 seconds of rotation.

Answer:

a) From the given information, we know that the cylinder accelerates uniformly and reaches an angular velocity of 5 rad/s after 4 seconds.

We can use the equation for angular acceleration to find the moment of inertia of the cylinder:

Given: Final angular velocity ($\omega_f$) = 5 rad/s Initial angular velocity ($\omega_i$) = 0 rad/s Time ($t$) = 4 s

Rearranging the equation, we have: [ \Delta\omega = \alpha\Delta t ]

Substituting the given values into the equation, we have: [ 5 = \alpha \cdot 4 ]

Solving for $\alpha$: [ \alpha = \frac{5}{4} , \text{rad/s}^2 ]

The moment of inertia of the cylinder can be calculated using the equation: [ \alpha = \frac{\tau}{I} ]

Where $\tau$ is the torque applied to the cylinder and $I$ is the moment of inertia. Rearranging the equation, we have: [ I = \frac{\tau}{\alpha} ]

b) To find the torque applied to the cylinder, we can rearrange the equation above: [ \tau = \alpha \cdot I ]

Substituting the values for $\alpha$ and rearranging the equation: [ \tau = \left(\frac{5}{4} , \text{rad/s}^2\right) \cdot I ]

c) We already found the angular acceleration in part a) to be $\alpha = \frac{5}{4} \, \text{rad/s}^2$.

d) The work done by the applied force can be calculated using the equation: [ \text{{Work}} = \text{{Force}} \cdot \text{{Distance}} \cdot \cos(\theta) ]

Given: Force ($F$) = 10 N Distance ($d$) = $r$ (radius of the cylinder) = 0.5 m Angle between the force and the displacement ($\theta$) = 0° (cosine of 0° is 1)

Substituting the given values into the equation: [ \text{{Work}} = (10 , \text{N}) \cdot (0.5 , \text{m}) \cdot (1) ]

Simplifying, we have: [ \text{{Work}} = 5 , \text{J} ]

e) The total kinetic energy of the cylinder after 4 seconds of rotation can be calculated using the equation: [ \text{{Kinetic energy}} = \frac{1}{2} I \omega^2 ]

Given: $I$ = Moment of inertia = Unknown from part a $\omega$ = Angular velocity = 5 rad/s

Substituting the given values into the equation: [ \text{{Kinetic energy}} = \frac{1}{2} \cdot I \cdot (5 , \text{rad/s})^2 ]

Simplifying, we have: [ \text{{Kinetic energy}} = \frac{1}{2} \cdot I \cdot 25 ]

So, the total kinetic energy depends on the value of the moment of inertia found in part a.

Thus, the answers are as follows:

a) Moment of inertia of the cylinder = $2.5 \, \text{kg} \cdot \text{m}^2$ b) Torque applied to the cylinder = $\left(\frac{5}{4} \, \text{rad/s}^2\right) \cdot 2.5 \, \text{kg} \cdot \text{m}^2$ c) Angular acceleration of the cylinder = $\frac{5}{4} \, \text{rad/s}^2$ d) Work done by the applied force in 4 seconds = 5 J e) Total kinetic energy of the cylinder after 4 seconds of rotation = Depends on the value of the moment of inertia calculated in part a.