A pendulum of mass m= 0.8 kg and length l=1 m is hanging from the ceiling. The massless string of the pendulum is attached at point P. The bob of the pendulum is a uniform shell (very thin hollow sphere) of radius r=0.4 m, and the length l of the pendulum is measured from the center of the bob. A spring with spring constant k= 7 N/m is attached to the bob (center). The spring is relaxed when the bob is at its lowest point (θ=0). In this problem, we can use the small-angle approximation sinθ≃θ and cosθ≃1. Note that the direction of the spring force on the pendulum is horizontal to a very good approximation for small angles θ. (See figure)

Take g= 10 m/s2

(a) Calculate the magnitude of the net torque on the pendulum with respect to the point P when θ=5∘. (magnitude; in Nm)

|τP|=

(b) What is the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘? (magnitude; in radians/s2)

|α|=

(c) What is the period of oscillation T of the pendulum? (in seconds)

T=

8:01 final exam question

|ôP|=1.308

|á|= 1.478
T= 1.526

To solve this problem, we need to use the principles of rotational dynamics for a simple pendulum. Let's go step by step to find the answers.

(a) To calculate the magnitude of the net torque, we need to find the torques contributed by each force acting on the pendulum: gravity and the spring force.

The torque due to gravity (τ_gravity) can be calculated using the formula τ_gravity = m * g * l * sin(θ), where m is the mass of the pendulum bob, g is the acceleration due to gravity, l is the length of the pendulum, and θ is the angle between the string and the vertical line.

In this problem, the small-angle approximation sinθ≃θ is given, so we can simplify the equation τ_gravity = m * g * l * θ.

The torque due to the spring force (τ_spring) can be calculated using the formula τ_spring = -k * x, where k is the spring constant and x is the displacement from the equilibrium position of the spring (which can be approximated as x = l * θ for small angles).

To find the net torque, we add the torques contributed by each force: τ_net = τ_gravity + τ_spring.

Substituting the formulas, τ_net = m * g * l * θ - k * (l * θ).

Now we can calculate the magnitude of the net torque when θ = 5∘:

|τP| = |τ_net| = |m * g * l * θ - k * (l * θ)|

Substituting the values given in the problem: m = 0.8 kg, g = 10 m/s^2, l = 1 m, k = 7 N/m, θ = 5∘:

|τP| = |0.8 * 10 * 1 * (5/180 * π) - 7 * (1 * (5/180 * π))|

Simplifying the expression gives |τP| = 0.4 * π - 7 * (5/180 * π) ≈ 1.375 Nm.

Therefore, the magnitude of the net torque on the pendulum is |τP| = 1.375 Nm.

(b) The angular acceleration (α) of the pendulum can be calculated using the formula α = θ¨, where θ¨ is the second derivative of the angular displacement with respect to time.

To find θ¨, we can use Newton's second law for rotational motion, τ = I * α, where τ is the net torque and I is the moment of inertia of the pendulum about point P.

For a uniform shell, the moment of inertia about an axis passing through its center of mass and perpendicular to its surface is I = (2/3) * m * r^2, where m is the mass of the shell and r is its radius.

Substituting the formulas, τ = (2/3) * m * r^2 * α.

Since we already know the magnitude of the net torque |τP| from part (a), we can substitute it into the equation: |τP| = (2/3) * m * r^2 * α.

Rearranging the equation to solve for α, α = |τP| / ((2/3) * m * r^2).

Substituting the given values: |α| = 1.375 / ((2/3) * 0.8 * (0.4)^2).

Simplifying the expression gives |α| ≈ 4.453 radians/s^2.

Therefore, the magnitude of the angular acceleration of the pendulum when θ = 5∘ is |α| ≈ 4.453 radians/s^2.

(c) The period of oscillation (T) of a simple pendulum can be calculated using the formula T = 2π * √(l / g), where l is the length of the pendulum and g is the acceleration due to gravity.

Substituting the given values: T = 2π * √(1 / 10).

Simplifying the expression gives T ≈ 2π * √(0.1) ≈ 2π * 0.316 ≈ 1.99 seconds.

Therefore, the period of oscillation of the pendulum is T ≈ 1.99 seconds.