Speed Calculation of a Hoop Rolling Up a Ramp

How to calculate the linear speed of a hoop rolling up a ramp?

A hoop is rolling without slipping along a horizontal surface with a forward speed of 5.99 m/s when it starts up a ramp that makes an angle of 18.70° with the horizontal. What is the linear speed of the hoop after it has rolled 1.66 m up as measured along the surface of the ramp?

Calculating the Linear Speed of the Hoop Rolling Up the Ramp

The linear speed of the hoop after it has rolled 1.66 m up the ramp is approximately 4.57 m/s.

To find the linear speed of the hoop after it has rolled 1.66 m up the ramp, we can use the principle of conservation of energy. The initial energy of the hoop is equal to the final energy of the hoop.

1. First, let's calculate the initial energy of the hoop. The initial energy consists of its translational kinetic energy and its rotational kinetic energy. The translational kinetic energy is given by: KE_trans = (1/2) * m * v^2, where m is the mass of the hoop and v is the initial linear speed of the hoop.

2. Next, let's calculate the rotational kinetic energy of the hoop. The rotational kinetic energy is given by: KE_rot = (1/2) * I * ω^2, where I is the moment of inertia of the hoop and ω is the angular speed of the hoop.

3. Since the hoop is rolling without slipping, the linear speed v is related to the angular speed ω by the equation: v = ω * R, where R is the radius of the hoop.

4. The moment of inertia of a hoop rotating about its central axis is given by: I = m * R^2.

5. Substituting the equations for KE_trans, KE_rot, and I into the conservation of energy equation, we get: (1/2) * m * v^2 + (1/2) * (m * R^2) * (v/R)^2 = m * g * h, where g is the acceleration due to gravity and h is the height the hoop has rolled up the ramp.

6. Now we can solve for the final linear speed v. Rearranging the equation, we get: v^2 + (v^2/R^2) = 2 * g * h. Simplifying further, we have: v^2 = (2 * g * h) / (1 + (1/R^2)).

7. Plugging in the given values, we have: v^2 = (2 * 9.8 m/s^2 * 1.66 m) / (1 + (1/1.66 m^(-2))). Solving this equation, we find v ≈ 4.57 m/s.

Therefore, the linear speed of the hoop after it has rolled 1.66 m up the ramp is approximately 4.57 m/s.