Question

A meter stick with a mass of 0.167 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. The 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 1.00 m, starting from rest.

Answer 1

Answer:

Part a)

change in potential energy is given as

$\Delta U = 0.82 J$

Part B)

angular speed of the rod is given as

$\omega = 5.42 rad/s$

Part c)

Linear speed of the end of the rod is given as

$v = 5.42 m/s$

Part d)

when a particle falls from rest to distance d = 1 m

$v = 4.42 m/s$

Explanation:

Part A)

As we know that the gravitational potential energy change is given as

$\Delta U = mgH$

$\Delta U = 0.167(9.81)(0.5)$

$\Delta U = 0.82 J$

Part B)

As we know that change in gravitational energy is equal to gain in kinetic energy

so we have

$\Delta U = \frac{1}{2}I\omega^2$

$0.82 = \frac{1}{2}(\frac{mL^2}{3})\omega^2$

$\omega^2 = \frac{6 \times 0.82}{0.167 (1)^2}$

$\omega = 5.42 rad/s$

Part c)

Linear speed of the end of the rod is given as

$v = L\omega$

$v = 5.42 m/s$

Part d)

when a particle falls from rest to distance d = 1 m

so we will have

$v = \sqrt{2gL}$

$v = \sqrt{2(9.81)(1)}$

$v = 4.42 m/s$

Answer 2

Following are the calculation to the given points:

Given:

$Mass(m) = 0.167\ kg\\\\length = 1\ m$

To find:

change in potential energy,  angular velocity, Linear velocity, and Compare the value with the linear velocity of the falling particle y = 1 m

Solution:

For point a:

When there is no friction, the mechanical energy is equal to the sum of the kinetic and potential energies.

The following is an example of a potential energy shift:

$\to \Delta U = U_f-U_o= -U_f$

Since this rod is a rigid body, external forces are exerted at its centre of mass. To calculate the energy, we'll use the linear density notion.

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

Let's subtitute.

$\to \Delta U= -gy(\mu dy)$        

We integrate

$\to U= - \mu g \int^{L}_{0} y\ dy \\\\ \to U= - \mu g \frac{L^2}{2}$

Let's use linear density as an alternative.

$\to U = - \frac{1}{2} mg L$

Let's calculate

    $\to \Delta U = - \frac{1}{2} \ 0.167 \ 9.8\ 1=-0.8183\ J$

For part b:

Energy is preserved via angular velocity.

$\to Emo = Emf$

The body spins and moves linearly.

$\to 0.8183 = \frac{1}{2} \ m \ v^2 + \frac{1}{2} I w^2$

The linear and angular variables are connected.

$\to v = w r$

we substitute

$\to 0.8183 = \frac{1}{2} m w^2 r^2 + \frac{1}{2} I w^2 = \frac{1}{2} w^2 (m L^2 + I)\\\\\to w=\sqrt{\farc{2 \times 0.8183 }{mL^2 +I}}$

At one extreme, the rotational inertia of a rod with such a rotating axis is:

$\to I = \frac{1}{3} m L^2$

Let's calculate

$\to I = \frac{1}{3}\times 0.167 \times 1^2= 0.0556\\\\\to w = \sqrt{\frac{2 \times 0.8183 }{0.167 \times 1^2+ 0.0556}}$

       $= \sqrt{\frac{1.6366 }{0.167+ 0.0556}}\\\\= \sqrt{\frac{1.6366 }{0.2226}}\\\\=7.352\ \frac{rad}{s}$

 For part c:

There is a connection between the angular and linear parameters.

$\to v = w \times r$

        $= 7.352 \times 1\\\\= 7.352 \ \frac{m}{s}$

 For part d:

We relate this to the fall of a body from h = 1.

$\to v^2 = v^2_0 - 2 g h$

$\to v = \sqrt{2gh}$

       $= \sqrt{2 \times 9.8 \times 1} \\\\ =4.427 \ \frac{m }{ s}$

We can observe that the body's speed is larger than the bolt's speed since some of the energy is consumed in the rotating movement.

Finally, using energy conservation, we can obtain the answers to the questions about the motion of the rod when rotated at its ends, which are:

Changes in potential energy= $\Delta U = -0.8183\ J$

Angular velocity = $w = 7.352 \ \frac{ rad}{ s}$

Linear velocity $=7.352 \ \frac{m}{ s}$

The speed of the bar falling out of the same height is larger than that of the speed of the revolving bar.

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