Now your bicycle wheel is rotating with an angular speed of 43.8 rad/s. The wheel has a mass of 4.9 kg and a radius of 48.6 cm. Treat the wheel as a solid thin disc.

(a)
Calculate the rotational kinetic energy of the wheel in J.
(b)
Calculate the angular speed the wheel must have if its rotational kinetic energy is doubled, in rad/s.

(a) To calculate the rotational kinetic energy of the wheel, we can use the formula:

Rotational kinetic energy (KE) = (1/2) * moment of inertia * angular speed^2

The moment of inertia of a solid thin disc is given by the formula:

moment of inertia (I) = (1/2) * mass * radius^2

Substituting the given values:

mass = 4.9 kg
radius = 48.6 cm = 0.486 m
angular speed = 43.8 rad/s

First, let's calculate the moment of inertia:

I = (1/2) * 4.9 kg * (0.486 m)^2
I = 0.57144 kg·m^2

Now, substitute the values of I and angular speed into the formula for rotational kinetic energy:

KE = (1/2) * 0.57144 kg·m^2 * (43.8 rad/s)^2
KE ≈ 543.26 J

Therefore, the rotational kinetic energy of the wheel is approximately 543.26 J.

(b) To calculate the new angular speed required to double the rotational kinetic energy, we can use the same formula for rotational kinetic energy:

Rotational kinetic energy (KE) = (1/2) * moment of inertia * (angular speed)^2

Since the rotational kinetic energy is doubled, we can write:

2KE = (1/2) * moment of inertia * (angular speed)^2

Divide both sides of the equation by (1/2) * moment of inertia:

2KE / ((1/2) * moment of inertia) = (angular speed)^2

Take the square root of both sides to isolate the angular speed:

√(2KE / ((1/2) * moment of inertia)) = angular speed

Substituting the values of KE and moment of inertia:

KE = 2 * 543.26 J ≈ 1086.52 J
moment of inertia = 0.57144 kg·m^2

√(2 * 1086.52 J / ((1/2) * 0.57144 kg·m^2)) = angular speed

Simplifying the equation:

√(2173.04 J / 0.57144 kg·m^2) = angular speed

√3800 ≈ angular speed

Therefore, the angular speed the wheel must have to double its rotational kinetic energy is approximately 61.62 rad/s.