A pendulum of mass m= 0.9 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= 14 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|=
unanswered
(b) What is the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘? (magnitude; in radians/s2)
|α|=
unanswered
(c) What is the period of oscillation T of the pendulum? (in seconds)
T=
torque=2.0071
angular acceleration=2.01516
lies!
To find the magnitude of the net torque on the pendulum, 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.
(a) The moment of inertia for a uniform thin shell of radius r is given by:
I = 2/3 * m * r^2
Plugging in the values for mass and radius:
I = 2/3 * 0.9 kg * (0.4 m)^2
= 2/3 * 0.9 kg * 0.16 m^2
= 0.096 kg * m^2
Now, we need to find the angular acceleration α. Using the small-angle approximation, we can relate the angular acceleration to the angular displacement θ:
α = g / l * θ
Plugging in the values for acceleration due to gravity and length:
α = 10 m/s^2 / 1 m * (5∘ * π/180) (converting degrees to radians)
= 10 m/s^2 * 0.0873 radians
= 0.873 radians/s^2
Now we can substitute the values of moment of inertia and angular acceleration into the torque equation:
τ = 0.096 kg * m^2 * 0.873 radians/s^2
= 0.084 Nm
Therefore, the magnitude of the net torque on the pendulum when θ = 5∘ is |τP| = 0.084 Nm.
(b) The magnitude of the angular acceleration α is already calculated as 0.873 radians/s^2.
(c) The period of oscillation T of a simple pendulum can be calculated using the formula:
T = 2π * √(l / g)
Substituting the values of length and acceleration due to gravity:
T = 2π *√(1 m / 10 m/s^2)
= 2π *√(0.1 s^2/m)
Simplifying further:
T = 2 * 3.14 * √(0.1)
= 2 * 3.14 * 0.316
T ≈ 1.99 seconds
Therefore, the period of oscillation of the pendulum is T = 1.99 seconds.