A 1-kg thin hoop with a 50-cm radius rolls down a 47° slope without slipping. If the hoop starts from rest at the top of the slope, what is its translational velocity after it rolls 16 m along the slope?

Question

A 1-kg thin hoop with a 50-cm radius rolls down a 47° slope without slipping. If the hoop starts from rest at the top of the slope, what is its translational velocity after it rolls 16 m along the slope?

Answer 1

To find the translational velocity of the hoop after it rolls 16 m along the slope, we need to consider its rotational and translational motion. Since the hoop is rolling without slipping, we can relate the rotational velocity to the translational velocity using the equation:

v = ω * r

where v is the translational velocity, ω is the rotational velocity, and r is the radius of the hoop.

Let's start by finding the initial angular velocity (ω_initial) of the hoop. Since the hoop starts from rest, its initial rotational velocity is 0 rad/s.

Next, we need to find the final angular velocity (ω_final) of the hoop. We can use the conservation of mechanical energy to relate the initial and final positions of the hoop. The potential energy at the top of the slope is converted into both rotational kinetic energy and translational kinetic energy:

m * g * h = (1/2) * m * v^2 + (1/2) * I * ω^2

where m is the mass of the hoop, g is the acceleration due to gravity, h is the vertical height of the slope, v is the translational velocity, and I is the moment of inertia of the hoop.

Since the hoop is a thin hoop, its moment of inertia is given by I = m * r^2, where r is the radius of the hoop.

At the top of the slope, the hoop is at a height h = r * sin(47°) above its final position after rolling 16 m.

Substituting these values into the equation, we can solve for ω_final:

m * g * r * sin(47°) = (1/2) * m * v^2 + (1/2) * m * r^2 * ω_final^2

Simplifying the equation, we have:

g * r * sin(47°) = (1/2) * v^2 + (1/2) * r^2 * ω_final^2

Now, let's solve for ω_final. Rearranging the equation, we get:

ω_final = sqrt((2 * g * r * sin(47°) - v^2) / r^2)

Substituting the given values, where g = 9.8 m/s^2, r = 0.5 m, and v = 16 m, we can calculate ω_final.

Once we have ω_final, we can use the equation v = ω_final * r to find the translational velocity v:

v = ω_final * r = sqrt((2 * g * r * sin(47°) - v^2) / r)

Now, we can plug in the values and calculate the translational velocity.

A 1-kg thin hoop with a 50-cm radius rolls down a 47° slope without slipping. If the hoop starts from rest at the top of the slope, what is its translational velocity after it rolls 16 m along the slope?

dc;k

To find the translational velocity of the hoop after it rolls 16 m along the slope, we need to consider its rotational and translational motion. Since the hoop is rolling without slipping, we can relate the rotational velocity to the translational velocity using the equation: