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= 12 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=
8:01 final question. The easiest one.
damon can u please tell me about the T that is the last part of this question
omega= sqrt((kl^2+mgl)/I)
please tell me the exact expression for this question.
|ôP|= 1.83
|á|= 1.84
T= 1.37
Can you plz give out the formula??
I = m L^2 + (2/3) m R^2
-(k L^2 + mgL)T = .885 d^2T/dt^2
well calculate the coefficient of T on the left e.g.
-(7 + 8) T = .885 d^2T/dt^2
.885 d^2T/dt^2 = - 15 T
Torque = -15T where T = 5 * pi/180
alpha = d^2T/dt^2 = - 15 T /.885
.885 w^2 = 15
w = 2 pi f = 2 pi/period = 4.11 for example and you solve fot period
To solve this problem, we need to analyze the forces and torques acting on the pendulum.
(a) To calculate the net torque on the pendulum with respect to the point P, we need to consider the torque due to the gravitational force and the torque due to the spring force.
The torque due to the gravitational force can be calculated as:
τ_grav = m * g * l * sin(θ)
where m is the mass of the pendulum, g is the acceleration due to gravity, l is the length of the pendulum, and θ is the angle of displacement.
The torque due to the spring force can be calculated as:
τ_spring = -k * x
where k is the spring constant and x is the displacement of the pendulum from its equilibrium position.
In this problem, the spring is relaxed when the pendulum is at θ=0, so the displacement x can be given as:
x = l * (1 - cos(θ))
where θ is the angle of displacement.
Now, let's substitute the given values into the equations and calculate |τP| when θ=5∘:
m = 0.9 kg
g = 10 m/s^2
l = 1 m
k = 12 N/m
θ = 5∘
First, calculate the displacement x:
x = l * (1 - cos(θ))
x = 1 * (1 - cos(5∘))
x = 0.0092 m
Then, calculate the torques:
τ_grav = m * g * l * sin(θ)
τ_grav = 0.9 * 10 * 1 * sin(5∘)
τ_grav = 0.0788 Nm
τ_spring = -k * x
τ_spring = -12 * 0.0092
τ_spring = -0.1104 Nm
The net torque is the sum of the two torques:
|τP| = |τ_grav| + |τ_spring|
|τP| = 0.0788 + 0.1104
|τP| = 0.1892 Nm
Therefore, the magnitude of the net torque on the pendulum with respect to the point P when θ=5∘ is 0.1892 Nm.
(b) To calculate the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘, we can use the equation of motion for a pendulum:
θ¨ = -g/l * sin(θ)
Substituting the given values:
g = 10 m/s^2
l = 1 m
θ = 5∘
θ¨ = -10/1 * sin(5∘)
θ¨ ≈ -0.0873 rad/s^2 (using small-angle approximation)
Therefore, the magnitude of the angular acceleration α=θ¨ of the pendulum when θ=5∘ is approximately 0.0873 rad/s^2.
(c) The period of oscillation T of a pendulum can be calculated using the equation:
T = 2π√(l/g)
Substituting the given values:
l = 1 m
g = 10 m/s^2
T = 2π√(1/10) ≈ 2.82 s
Therefore, the period of oscillation T of the pendulum is approximately 2.82 seconds.