Determining Angular Momentum of a Rolling Hoop
What is the correct statement concerning the angular momentum of a 2.0 kg rolling hoop on a horizontal surface?
1. The angular momentum is nonzero because the hoop is rotating.
2. The angular momentum is zero because the hoop is rolling without slipping.
3. The angular momentum is maximum because the hoop's linear speed is constant.
4. The angular momentum is negative because the hoop is moving to the right.
Correct Statement:
The correct statement is that the angular momentum is nonzero because the hoop is rotating. Hence, option 1 is correct.
Given that a hoop with a mass of 2.0 kg is rolling without slipping on a horizontal surface with a constant linear speed of 6.0 m/s to the right, we need to determine the correct statement regarding the angular momentum of the hoop.
In this case, the linear speed of the center of mass (v) is related to the angular speed (ω) by the formula: v = R * ω, where R is the radius of the hoop.
To find the angular speed, rearrange the formula to ω = v / R. Calculate the radius of the hoop as R.
The angular momentum (L) of the hoop is given by the formula: L = I * ω, where I is the moment of inertia of the hoop. For a hoop rotating about its axis, the moment of inertia (I) is given by I = m * R², where m is the mass of the hoop.
Substitute the values into the formula to calculate the angular momentum (L) using the equation: L = 2.0 * v * R. Since the linear speed of the center of mass (v) is constant, the angular momentum is nonzero because the hoop is rotating about its axis.
Therefore, the correct statement is that the angular momentum is nonzero because the hoop is rotating (option 1).