In the figure below, a constant horizontal force Fapp of magnitude 17 N is applied to a wheel of mass 8 kg and radius 0.70 m. The wheel rolls smoothly on the horizontal surface, and the acceleration of its center of mass has magnitude 0.75 m/s2.
(a) In unit-vector notation, what is the frictional force on the wheel?
1 N
(b) What is the rotational inertia of the wheel about the rotation axis through its center of mass?
2 kg·m2
To find the frictional force on the wheel, we can use Newton's second law. The equation is given by:
F_net = ma
Where F_net is the net force acting on the wheel, m is the mass of the wheel, and a is the acceleration of the center of mass.
In this case, the only horizontal force acting on the wheel is the applied force Fapp, and the only vertical force is the weight of the wheel. The frictional force acts in the opposite direction to the applied force and prevents the wheel from sliding.
Since the wheel is rolling smoothly, there is no slipping, and the frictional force can be expressed as:
Frictional force = Fapp - Weight
The weight of the wheel can be calculated using the formula:
Weight = mg
Where m is the mass of the wheel and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, let's solve for the frictional force:
1. Calculate the weight of the wheel:
Weight = (mass) x (acceleration due to gravity)
Weight = (8 kg) x (9.8 m/s^2)
Weight = 78.4 N
2. Substituting the values into the frictional force equation:
Frictional force = Fapp - Weight
Frictional force = 17 N - 78.4 N
Therefore, the frictional force on the wheel is equal to -61.4 N (opposite direction to the applied force).
Now let's move on to part (b) where we need to find the rotational inertia of the wheel about the rotation axis through its center of mass.
The rotational inertia (also known as the moment of inertia) depends on the mass distribution of an object about the axis of rotation. For a solid disc or wheel, the rotational inertia (I) is given by the formula:
I = (1/2) * m * r^2
Where m is the mass of the wheel and r is the radius of the wheel.
Let's substitute the given values into the formula to find the rotational inertia of the wheel:
I = (1/2) * (8 kg) * (0.70 m)^2
I = (1/2) * 8 kg * 0.49 m^2
I = 2 kg·m^2
Therefore, the rotational inertia of the wheel about the rotation axis through its center of mass is 2 kg·m^2.