Physics, 21.03.2020 04:42 whityelaine7997

A meter stick with a mass of 0.155 kg is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released. As it swings through the vertical, calculate:
a. change in gravitational potential energy that has occurred;
b. the angular speed of the stick;
c. the linear speed of the end of the stick opposite the axis.
d. Compare the answer in part (c) to the speed of a particle that has fallen 1.00m, starting from rest.

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The energy conservation allows to find the results for the questions about the movement of the rod when rotating at its ends are:

a) The change in potential energy is: DU = -0.76 J

b) The angular velocity is: w = 2.71 rad / s

c) The linear velocity is: 2.71 m / s

d) The speed of the bar falling from the same height is greater than the speed of the rotating bar.

Given parameters.

Rod mass m = 0.155 kg Rod length m = 1 m.

To find.

a) The potential energy change.

b) The angular velocity.

c) Linear velocity.

d) Compare with the linear velocity of a falling particle y = 1 m

A) The mechanical energy is the sum of the kinetic energy and the potential energies, if there is no friction this magnitude is conserved.

The potential energy change is:

ΔU = U_f - U_o

ΔU = - U_f

The rod is a rigid body therefore the external force is applied at its center of mass, let's use the concept of linear density to find the energy, see attached.

$\mu = \frac{dm}{dy}$

dm = μ dy

Let's subtitute.

$\Delta U = - g y (\mu dy)$

We integrate

$U = - \mu g \int\limits^L_0 {y} \, dy$

$U = - \mu g \frac{L^2}{2}$

Let's substitute with the linear density

U = $- \frac{1}{2} m g L$

Let's calculate

ΔU = - ½ 0.155 9.8 1

ΔU = -0.76 J

B) Angular velocity

Energy is conserved

Emo = Emf

The body rotates and moves linearly.

0.76 = ½ m v² + ½ I w²

The linear and angular variables are related.

v = w r

we substitute

0.76 = ½ m w² r² + ½ I w² = ½ w² (m L² + I)

$w= \sqrt{\frac{2 \ 0.76}{mL^2 + I } }$

The moment of inertia of a rod with an axis of rotation is one extreme is:

I = ⅓ mL²

Let's calculate

I = ⅓ 0.155 1²

I = 0.05166

w = $\sqrt{\frac{2 \ 0.76 }{0.155 \ 1^2 + 0.051667} } = 7.3548$

w = 2.71 rad / s

C) The angular and linear variables are related.

v = w r

v = 2.71 1

v = 2.71 m / s

d) We compare with the fall of a body from h = 1

v² = v₀² - 2 g h

v = $\sqrt{2gh}$

Let's calculate

v = $\sqrt{2\ 9.8 \ 1}$

v = 4.427 m / s

We can see that the speed of the body is greater than the speed of the rod, this is because part of the energy is used in the rotational movement.

In conclusion, using the energy conservation we can find the results for the questions about the motion of the rod when rotated at its ends are:

a) The change in potential energy is: DU = -0.76 J

b) The angular velocity is: w = 2.71 rad / s

c) The linear velocity is: 2.71 m / s

d) The speed of the bar falling from the same height is greater than the speed of the rotating bar.

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A meter stick with a mass of 0.155 kg is pivoted about one end so it can rotate without friction abo

Answer from: 240662

a) P.E = - 0.76J

b) W = 2.7 rad/s

c) V = 2.7m/a

d) V = 4.43 m/s

Explanation: Please find the attached files for the solution

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