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= 9 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=
To solve this problem, we need to analyze the forces and torques acting on the pendulum at a given angle θ. Let's go step by step:
(a) To calculate the magnitude of the net torque on the pendulum with respect to point P when θ = 5°, we need to consider the torque due to gravity and the torque due to the spring force.
The torque due to gravity τg is given by the equation:
τg = m * g * l * sin(θ)
where m is the mass of the bob, g is the acceleration due to gravity, l is the length of the pendulum, and θ is the angle.
In this problem, we are using the small-angle approximation sinθ ≈ θ. So, we can substitute sin(θ) with θ in the equation:
τg = m * g * l * θ
The torque due to the spring force τs is given by the equation:
τs = -k * x
where k is the spring constant and x is the displacement of the spring from its equilibrium position. At θ = 5°, the displacement x can be approximated as:
x = l * θ
So, we substitute x with l * θ in the equation:
τs = -k * l * θ
The net torque τP is the sum of τg and τs:
τP = τg + τs
= m * g * l * θ - k * l * θ
Now, substitute the given values: m = 0.8 kg, g = 10 m/s^2, l = 1 m, k = 9 N/m, and θ = 5° (convert to radians):
τP = 0.8 kg * 10 m/s^2 * 1 m * (5° * π/180)
- 9 N/m * 1 m * (5° * π/180)
Calculating the above expression gives:
|τP| ≈ 0.329 Nm
So, the magnitude of the net torque on the pendulum is approximately 0.329 Nm.
(b) To find the magnitude of the angular acceleration α (θ¨) of the pendulum when θ = 5°, we need to use Newton's second law for rotational motion:
τP = I * α
where I is the moment of inertia of the pendulum and α is the angular acceleration.
For a hollow sphere, the moment of inertia is given by the equation:
I = 2/3 * m * r^2
where m is the mass of the bob and r is the radius of the bob.
Substituting the given values: m = 0.8 kg and r = 0.4 m, we can calculate:
I = 2/3 * 0.8 kg * (0.4 m)^2
Now, substitute the calculated value of τP (0.329 Nm) into the equation τP = I * α:
0.329 Nm = [2/3 * 0.8 kg * (0.4 m)^2] * α
Solving for α gives:
|α| ≈ 0.371 rad/s^2
So, the magnitude of the angular acceleration of the pendulum is approximately 0.371 radians/s^2.
(c) The period of oscillation T of a simple pendulum can be calculated using the equation:
T = 2π * sqrt(l / g)
where l is the length of the pendulum and g is the acceleration due to gravity.
Substituting the given value l = 1 m and g = 10 m/s^2, we can calculate:
T = 2π * sqrt(1 m / 10 m/s^2)
Calculating the above expression gives:
T ≈ 2.512 s
So, the period of oscillation of the pendulum is approximately 2.512 seconds.