Pendulum Physics Question

Question

A 3.6 kg pendulum bob hangs on the end of a cord 4.0 m long. The pendulum bob is raised so that the cord makes a 25° angle with the vertical.The bob is released from rest.

(a) What is the gravitational potential energy of the pendulum bob at the instant it is released? Take the lowest position in the swing as the reference level where potential energy is zero.

(b) What is the speed of the bob at the lowest point in its swing?

(c) What is the tension in the cord when the bob is at the lowest point in its swing?

Answer 1

(a)

At the point of release, the height of the bob is

L - Lcosθ = L(1 - cosθ)

Potential energy is

U = mgL(1 - cosθ)

You know m, g, L, and θ. Plug in the numbers and get U in joules.

(b)

Set the result from part (a) equal to kinetic energy, and solve for the speed.

mgL(1 - cosθ) = mv²/2

v = √[2gL(1 - cosθ)]

Plug in the numbers and get v in m/s.

(c)

T = tension

Since the motion is along a circular arc, you can set the net radial force equal to the centripetal force and solve for the tension. At the lowest point,

T - mg = mv²/L ⇒

T = mg + mv²/L

Plug in the numbers and get T in newtons.

Answer 2

Always draw a picture.....

With a cord length of 4 m, and a 25 degree angle from vertical, the length from the pivot is 4*cos 25 = 3.63 m. This means the bob is 0.37 m above the reference plane.

Potential energy (PE) = F * d, where F is m*g (g is the acceleration of gravity: 9.81 m/sec^2) d from above is 0.37

So PE = 3.6 * 9.81 * 0.37 = 13.1

At the bottom of the swing, the Potential Energy is converted to Kinetic Energy (KE) = 1/2 * m * v^2

Plug in the numbers: 1/2 * 3.6 * v^2 = 13.1

v^2 = 26.2 / 3.6, v = 2.70

Finally. the tension in the cord is the sum of the weight of the bob and the centripetal force**. F = m * v^2 / r

(r is the length of the cord, which the radius of the circular motion)

F = 3.6 * (2.7)^2 / 4 = 6.55

So the tension in the cord is 6.55 + 3.6 * 9.81 = 41.9 N