A solid cylinder with a radius of 0.2 meters and a mass of 5 kg is released from rest and rolls down an inclined plane with a coefficient of kinetic friction of 0.1. Determine the angular acceleration of the cylinder.
Guide On Rating System
Vote
To determine the angular acceleration of the cylinder, we need to find the net torque acting on it.
The net torque is given by the equation:
τ_net = Iα
where τ_net is the net torque, I is the moment of inertia of the cylinder, and α is the angular acceleration.
The moment of inertia of a solid cylinder rolling without slipping about its axis is given by the equation:
I = (1/2)MR^2
where M is the mass of the cylinder and R is the radius of the cylinder.
In this case, M = 5 kg and R = 0.2 m. Substituting these values into the equation, we get:
I = (1/2)(5 kg)(0.2 m)^2 = 0.1 kg·m^2
Next, we need to find the torque due to friction. The torque due to friction is given by the equation:
τ_friction = μmgR
where μ is the coefficient of kinetic friction, m is the mass of the cylinder, g is the acceleration due to gravity, and R is the radius of the cylinder.
In this case, μ = 0.1, m = 5 kg, g ≈ 9.8 m/s^2, and R = 0.2 m. Substituting these values into the equation, we get:
τ_friction = (0.1)(5 kg)(9.8 m/s^2)(0.2 m) = 0.98 N·m
Since the cylinder is rolling without slipping, the angular acceleration is related to the linear acceleration by the equation:
α = a/R
where α is the angular acceleration, a is the linear acceleration, and R is the radius of the cylinder.
The linear acceleration can be determined using Newton's second law applied to the translational motion of the cylinder:
F_net = ma
where F_net is the net force, m is the mass of the cylinder, and a is the linear acceleration.
In this case, the net force acting on the cylinder is the force of gravity (weight) minus the force of friction:
F_net = mg - f_friction
where m is the mass of the cylinder, g is the acceleration due to gravity, and f_friction is the force of friction.
The force of gravity is given by:
mg = (5 kg)(9.8 m/s^2) = 49 N
The force of friction is given by:
f_friction = μmg = (0.1)(5 kg)(9.8 m/s^2) = 4.9 N
Substituting these values into the equation for F_net, we get:
F_net = 49 N - 4.9 N = 44.1 N
Finally, using Newton's second law, we can solve for the linear acceleration:
a = F_net / m = 44.1 N / 5 kg = 8.82 m/s^2
Substituting this value into the equation for α, we get:
α = a / R = 8.82 m/s^2 / 0.2 m = 44.1 rad/s^2
Therefore, the angular acceleration of the cylinder is 44.1 rad/s^2.
Got a question with my answer?
Cheapmath.com is completely free!
Asking question is free.
Getting answers is free.
+99