Angular Acceleration Calculation on a Lug Nut

What is the angular acceleration of a 75 g lug nut when a lug wrench applies a 135 N-m torque to it?

What is the tangential acceleration at the outer surface?

What factor was not considered which causes this acceleration to be so large?

Answer:

The angular acceleration of the 75 g lug nut when a 135 N-m torque is applied to it is 2.18 × 10^7 rad/s^2.

The tangential acceleration at the outer surface is 2.18 × 10^5 m/s^2.

The factor not considered that causes the acceleration to be so large is the small radius of the lug nut.

To determine the angular acceleration of the lug nut, we can use the torque formula: Torque (τ) = Moment of inertia (I) * Angular acceleration (α).

The moment of inertia of the hollow cylinder can be calculated using the formula: I = (1/2) * m * (r1^2 + r2^2), where m is the mass and r1 and r2 are the inner and outer radii.

Given: Mass of the lug nut (m) = 75 g = 0.075 kg, Inner radius (r1) = 0.85 cm = 0.0085 m, Outer radius (r2) = 1.0 cm = 0.01 m, Torque (τ) = 135 N-m.

Calculating the moment of inertia: I = (1/2) * 0.075 * (0.0085^2 + 0.01^2) = 6.19 × 10^-6 kg·m^2.

Now we can solve for the angular acceleration (α): τ = I * α, 135 = 6.19 × 10^-6 * α, α = 135 / (6.19 × 10^-6) = 2.18 × 10^7 rad/s^2.

To find the tangential acceleration at the outer surface, we use the formula: Tangential acceleration (at) = Radius (r) * Angular acceleration (α).

Using the outer radius (r2) = 0.01 m: at = 0.01 * 2.18 × 10^7 = 2.18 × 10^5 m/s^2.

The small radius of the lug nut contributes to the large tangential acceleration because the acceleration is directly proportional to the radius.