Question

A thin rod of length 0.75 m and mass 0.42 kg 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 an angular speed of 4.0 rad/s. neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position, (b) and how far above that position the center of mass rises.

Answer 1

The rod's kinetic energy at its lowest position is 1.26 J, calculated using the formula for rotational kinetic energy. The center of mass of the rod rises to a height of approximately 30.6 cm above the lowest point, using the principle of conservation of mechanical energy to equate potential and kinetic energies.

Step-by-step explanation:

A thin rod acting as a pendulum is an example of physical principles related to rotational motion and conservation of energy. To find the rod's kinetic energy at its lowest position, we use the formula for rotational kinetic energy, K = (1/2)Iω2, where 'I' is the moment of inertia and 'ω' is the angular speed. For a rod rotating about one end, the moment of inertia, 'I', is given by (1/3)ml2, with 'm' being the mass of the rod and 'l' its length.

We can find the kinetic energy of the rod at its lowest position, where the angular speed is 4.0 rad/s: K = (1/2)Iω2 = (1/2)(1/3)(0.42 kg)(0.75 m)2(4.0 rad/s)2. After calculation, this gives K = 1.26 J (Joules).

To determine how far above the lowest position the center of mass rises, we have to consider the conservation of mechanical energy. At the highest point, all kinetic energy is converted into potential energy (assuming no energy losses), so we have mgh = K, where 'g' is the acceleration due to gravity, 'h' is the height gained, and 'm' is the mass of the rod. Therefore, h = K / (mg). Plugging in the values, we get h = 1.26 J / (0.42kg * 9.81 m/s2) = 0.306 m, or approximately 30.6 cm above the lowest position.

Answer 2

(A) 0.63 J

(B) 0.15 m

Step-by-step explanation:

length (L) = 0.75 m

mass (m) =0.42 kg

angular speed (ω) = 4 rad/s

To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)

I = Ic + m
`h^(2)`

Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis

h is the horizontal distance between the center of mass and the rotational axis of the rod

I =
`((1)/(12))(mL^(2) ) + m([tex](L)/(2)`)^{2}[/tex]

I =
`((1)/(12))(0.42 x 0.75^(2) ) + ( 0.42 x ([tex](0.75)/(2)`)^{2}[/tex])

I = 0.07875 kg.m^{2}

(A) rods kinetic energy = 0.5I
`ω^(2)`

= 0.5 x 0.07875 x
`4^(2)` = 0.63 J 0.15 m

(B) from the conservation of energy

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

Ki + Ui = Kf + Uf

at the maximum height velocity = 0 therefore final kinetic energy = 0

Ki + Ui = Uf

Ki = Uf - Ui

Ki = mg(H-h)

where (H-h) = rise in the center of mass

0.63 = 0.42 x 9.8 x (H-h)

(H-h) = 0.15 m