Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
To solve this problem, we can use the equation for torque:
τ = r * F * sin(θ)
where τ is the torque, r is the radius, F is the force, and θ is the angle between the force and the lever arm.
(a) To find the torque exerted:
τ = 0.280 m * 180 N * sin(90°)
Since sin(90°) is equal to 1, we have:
τ = 0.280 m * 180 N
Calculating this, we get:
τ = 50.4 N·m
So the torque exerted is 50.4 N·m.
(b) To find the angular acceleration assuming negligible opposing friction, we can use the equation:
τ = I * α
where τ is the torque, I is the moment of inertia, and α is the angular acceleration.
Rearranging the equation, we have:
α = τ / I
The moment of inertia for a solid disk is given by:
I = (1/2) * M * R^2
where M is the mass and R is the radius of the disk.
Substituting the given values, we have:
I = (1/2) * 75.0 kg * (0.280 m)^2
Calculating this, we find:
I = 2.94 kg·m^2
Now, we can find the angular acceleration:
α = 50.4 N·m / 2.94 kg·m^2
Calculating this, we get:
α ≈ 17.14 rad/s^2
Therefore, the angular acceleration, assuming negligible opposing friction, is approximately 17.14 rad/s^2.
(c) To find the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis, we need to calculate the torque due to friction first.
τ_f = r_f * F_f * sin(θ_f)
where τ_f is the torque due to friction, r_f is the distance from the axis to the point of application of the frictional force, F_f is the magnitude of the frictional force, and θ_f is the angle between the lever arm and the direction of the frictional force.
Converting the given distance to meters, we have:
r_f = 0.015 m
Substituting the values, we get:
τ_f = 0.015 m * 20.0 N * sin(90°)
Since sin(90°) is equal to 1, we have:
τ_f = 0.015 m * 20.0 N
Calculating this, we find:
τ_f = 0.3 N·m
Now, we can find the total torque:
τ_total = τ - τ_f
Substituting the given torque and the torque du etiquette to friction, we get:
τ_total = 50.4 N·m - 0.3 N·m
Calculating this, we find:
τ_total = 50.1 N·m
Using the same equation as before, we can find the angular acceleration:
α = τ_total / I
Substituting the values, we have:
α = 50.1 N·m / 2.94 kg·m^2
Calculating this, we get:
α ≈ 17.03 rad/s^2
Therefore, the angular acceleration, considering the opposing frictional force, is approximately 17.03 rad/s^2.
To find the torque exerted on the grindstone, we can use the equation:
Torque (τ) = Force (F) * Distance (r)
(a) Substituting the given values, we have:
τ = 180 N * 0.280 m
τ = 50.4 N·m
The torque exerted on the grindstone is 50.4 N·m.
To find the angular acceleration, we can use the equation:
Torque (τ) = Moment of Inertia (I) * Angular Acceleration (α)
(b) Since the grindstone is a solid disk, the moment of inertia (I) can be calculated using the formula:
I = (1/2) * m * r^2
Substituting the given values, we have:
I = (1/2) * 75.0 kg * (0.280 m)^2
I = 1.47 kg·m^2
Now, we can rearrange the torque equation to solve for the angular acceleration:
α = τ / I
α = 50.4 N·m / 1.47 kg·m^2
α ≈ 34.42 rad/s^2
The angular acceleration assuming negligible opposing friction is approximately 34.42 rad/s^2.
(c) When considering the opposing frictional force, we need to account for its torque as well. The torque due to friction can be calculated as:
τ_friction = Frictional Force (F_friction) * Distance (r_friction)
Substituting the given values, we have:
τ_friction = 20.0 N * 0.015 m
τ_friction = 0.3 N·m
To find the net torque, we subtract the torque due to friction from the original torque:
τ_net = τ - τ_friction
τ_net = 50.4 N·m - 0.3 N·m
τ_net = 50.1 N·m
Now, we can calculate the angular acceleration accounting for friction:
α = τ_net / I
α = 50.1 N·m / 1.47 kg·m^2
α ≈ 34.09 rad/s^2
The angular acceleration considering the opposing frictional force is approximately 34.09 rad/s^2.
sdgfgdfsd 2
τ=FR= 180•0.28 = 50.4 N•m
I=mR²/2 = 75•0.28²/2=2.94 N•m²
τ=Iε =>
ε= τ/I = 50.4/2.94 = 17.14 rad/s²
τ₁=FR-F(fr)r =180•0.28 - 20•0.015 = 50.4 – 0.3 = 50.1 N•m
ε₁= τ₁/I = 50.1/2.94 = 17.04 rad/s²