Question

A thin uniform rod (mass = 0.53 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of 1.4 s and an angular amplitude of 3.9°.

(a) What is the length of the rod?
(b) What is the maximum kinetic energy of the rod as it swings?

Answer 1

To find the length of the rod, use the formula T = 2π√(L/g), where T is the period and g is the acceleration due to gravity. The maximum kinetic energy of the rod can be found using the formula KE = (1/2)Iω², where I is the moment of inertia and ω is the angular frequency.

Step-by-step explanation:

To determine the length of the rod, we can use the formula for the period of a simple pendulum:

T = 2π√(L/g)

Where T is the period, L is the length of the rod, and g is the acceleration due to gravity.

Given the period (T) as 1.4 s and the angular amplitude as 3.9°, we can convert the amplitude to radians by multiplying by π/180. So, α = 3.9° × (π/180) = 0.068 radians.

Next, we can calculate the angular frequency (ω) using the formula ω = 2π/T.

Once we have the angular frequency, we can find the length (L) of the rod by rearranging the formula: L = (g/ω²).

The maximum kinetic energy of the rod can be calculated using the formula for rotational kinetic energy:

KE = (1/2)Iω², where I is the moment of inertia. For a thin rod rotating about one end, the moment of inertia is (1/3)mL², where m is the mass of the rod and L is its length.

Answer 2

(a) L = 0·73 m

(b) 4·39 ×
`10^(-3)` J

Step-by-step explanation:

(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment

Let L m be the length of the rod and β be the angle between the rod and the vertical

Let α be the angular acceleration of the rod

As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ

Assuming that the value of amplitude of this oscillation to be small

As torque = moment of inertia × angular acceleration

0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)

For small oscillations, α = ω² × β

After substituting the value of α and solving we get

ω = √((3 × g) ÷ (2 × L))

Time period = (2 × π) ÷ ω = (2 × π) ÷ √((3 × g) ÷ (2 × L))

∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4

Substituting the value of g as 9·8 m/s² and solving we get

L = 0·73 m

(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving

∴ Taking the reference for finding the potential energy as the lowest point

Maximum potential energy = Maximum kinetic energy

As total energy is constant, since there is no dissipative force

Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2

∴ Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get

Maximum potential energy = 4·39 ×
`10^(-3)` J

∴ Maximum kinetic energy = 4·39 ×
`10^(-3)` J