A thin rod of length 1.3 m and mass 88 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 2.39 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

Explanation:

Given:

Length, L = 1.3 m

Mass, M = 88 g

= 0.088 kg

Angular velocity, w = 2.39 rad/s

Inertia of the rod = inertia of a cylinder = 1/12 M × L^2

Rotational inertia, I = Icm + I

= 1/12 M × L^2 + M × L/2

= 0.088 × (1/12 × 1.3^2 + 1.3/2)

= 0.088 × 0.79

= 0.05 kgm^2

Kinetic energy, E = 1/2 × l × w^2

= 1/2 × 0.05 × 2.39^2

= 0.142 J

B.

Using conservation of total mechanical energy,

Potential energy = kinetic energy

M × g × h = 0.142

h = 0.142/(9.8 × 0.088)

= 0.164 m

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

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