A bicycle tire has a mass of 2.54 kg and a radius of 0.346 m.
(a) Treating the tire as a hoop, what is its moment of inertia about an axis passing through the hub at its center?
kg · m2
(b) What torque is required to produce an angular acceleration of 0.710 rad/s2?
N · m
(c) What friction force applied tangentially to the edge of the tire will create a torque of that magnitude?
N
To find the answers to these questions, we need to use the relevant formulas and equations related to moment of inertia, torque, and friction force.
(a) Moment of Inertia (I) of a hoop can be calculated using the formula:
I = m * r^2
where m is the mass of the hoop and r is the radius.
Substituting the given values, we have:
I = 2.54 kg * (0.346 m)^2
I = 0.316 kg · m^2
So, the moment of inertia of the bicycle tire about an axis passing through the hub at its center is 0.316 kg · m^2.
(b) Torque (τ) can be calculated using the formula:
τ = I * α
where I is the moment of inertia and α is the angular acceleration.
Rearranging the formula to solve for torque, we have:
τ = I * α
τ = 0.316 kg · m^2 * 0.710 rad/s^2
τ = 0.224 N · m
So, the torque required to produce an angular acceleration of 0.710 rad/s^2 is 0.224 N · m.
(c) Friction force (F) can be calculated using the formula:
τ = F * r
where τ is the torque, F is the friction force, and r is the radius.
Rearranging the formula to solve for friction force, we have:
F = τ / r
F = 0.224 N · m / 0.346 m
F = 0.647 N
So, the friction force applied tangentially to the edge of the tire that will create a torque of that magnitude is 0.647 N.
(a) The moment of inertia of a hoop about an axis passing through its center is given by the formula:
I = m * r^2
where I is the moment of inertia, m is the mass of the hoop, and r is the radius of the hoop.
Substituting the given values, we have:
I = 2.54 kg * (0.346 m)^2
Calculating this, we find:
I ≈ 0.977 kg · m^2
So the moment of inertia of the tire about an axis passing through the hub at its center is approximately 0.977 kg · m^2.
(b) The torque required to produce an angular acceleration can be calculated using the formula:
T = I * α
where T is the torque, I is the moment of inertia, and α is the angular acceleration.
Substituting the given values, we have:
T = 0.977 kg · m^2 * 0.710 rad/s^2
Calculating this, we find:
T ≈ 0.693 N · m
So the torque required to produce an angular acceleration of 0.710 rad/s^2 is approximately 0.693 N · m.
(c) The torque created by a force applied tangentially to the edge of the tire can be calculated using the formula:
T = r * F
where T is the torque, r is the radius of the tire, and F is the applied force.
Rearranging the formula, we find:
F = T / r
Substituting the given values, we have:
F = 0.693 N · m / 0.346 m
Calculating this, we find:
F ≈ 2 N
So the friction force applied tangentially to the edge of the tire that will create a torque of that magnitude is approximately 2 N.