(b) Given that Icm = ½MR2 for a uniform solid wheel,

(i) show that the rotational kinetic energy is half the translational kinetic energy for such a wheel.
(ii) For the wheel of diameter 10 cm, find its final speed after traversing a distance of 10 m down an incline of 25º. Assume it starts from rest.
(iii) How long does it take to do this?
(iv) Which will travel faster down the slope: A uniform solid wheel or a hoop?
(v) Evaluate the time for the solid hoop to roll down the 10 m.

You need to show some effort of your own here. here is the way to proceed.

i) Use the facts that V = R w
and I = (1/2) M R^2
Compare (1/2) R V^2 with (1/2) I w^2, after substituting for I and w.

ii) Apply conservation of energy, making sure you count both translational and rotational KE

iii) Compute final speed at the bottom using the final kinetic energy. The average speed is half that. T = distance travelled/(average speed)

iv) Use same approach, but the appropriate moment of inertia formula

v) Same apprach as iii)

ii) Apply conservation of energy, making sure you count both translational and rotational KE

now with this part Im not sure what you mean?
i understand the conservation of energy but that wont give me the speed of the wheel

To answer these questions, you'll need to use a few concepts from rotational kinematics and dynamics. Let's go step by step to find the answers:

(i) To show that the rotational kinetic energy is half the translational kinetic energy for a uniform solid wheel, we can use the formulas for rotational and translational kinetic energy.

Translational kinetic energy (Kt) is given by Kt = (1/2) * mv^2, where m is the mass of the wheel and v is its linear speed.

Rotational kinetic energy (Kr) is given by Kr = (1/2) * Iω^2, where I is the moment of inertia of the wheel and ω is its angular velocity.

For a uniform solid wheel, the moment of inertia is given by I = (1/2)MR^2, where M is the mass of the wheel and R is its radius.

By substituting the values of I and ω into the formula for rotational kinetic energy, we get Kr = (1/2) * (1/2)MR^2 * (v/R)^2 = (1/4)Mv^2.

This means the rotational kinetic energy is indeed half the translational kinetic energy for a uniform solid wheel.

(ii) To find the final speed of the wheel after traversing a distance of 10 m down an incline of 25º, we can use the conservation of mechanical energy.

The initial mechanical energy of the wheel is given by the sum of its potential energy (mgh, where h is the initial height) and its rotational kinetic energy (Kr = (1/4)Mv^2, as shown in part (i)).

The final mechanical energy of the wheel is the sum of its translational kinetic energy (Kt = (1/2)mv^2) and potential energy at the final position (mgh', where h' is the final height at the bottom of the incline).

Since the wheel starts from rest, its initial kinetic energy is zero. Therefore, we have:

mgh + (1/4)Mv^2 = (1/2)mv^2 + mgh'

Here, we need to rearrange the equation to solve for v.

(iii) To find the time it takes for the wheel to traverse the given distance, we can use the kinematic equation relating distance, initial velocity, time, and acceleration.

Since the wheel starts from rest, its initial velocity (u) is zero. The acceleration (a) can be calculated using the component of the gravitational force acting down the incline (a = g*sin(θ), where θ is the angle of the incline).

The distance (d) is given as 10 m.

Using the kinematic equation d = ut + (1/2)at^2, we can solve for t.

(iv) To determine which will travel faster down the slope, a uniform solid wheel or a hoop, we need to compare their rotational inertias.

For a uniform solid wheel, the moment of inertia is I = (1/2)MR^2, as mentioned before.

For a hoop (a thin-walled cylinder), the moment of inertia is I = MR^2.

Comparing the two, we see that the moment of inertia for the hoop is greater than that of the uniform solid wheel. Therefore, the hoop will be slower, and the uniform solid wheel will travel faster down the slope.

(v) To evaluate the time for the solid hoop to roll down the 10 m, you can follow the same procedure as in part (iii), using the moment of inertia for a hoop (I = MR^2) and solving for time.