Home Page Textbooks Physics: Principles and Problems Page 200: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 200: Section Review

Exercise 5

Solution 1

Solution 2

Step 1

1 of 4

We need to calculate angular displacement of the hour hand and the minute hand over a period of 2 hours.

Step 2

2 of 4

During that period, the hour hand will move for the two out of twelve hours on the clock, That meanst that the number of revolutions an hour hand will make is:

$$
n_h=-frac{2}{12}
$$

During the same period, the minute hand will move for 120 minutes which fits two whole revolutions:

$$
n_m=-2
$$

Both are negative because the rotation is clockwise, which is considered negative direction in mathematics.

Step 3

3 of 4

Now we can calculate the angular displacement of hour hand as:

$$
theta_h=n_hcdot 2cdot pi=frac{2}{12}cdot 2cdot pi=-1.05,,rm{rad}
$$

And angular displacement of minute hand as:

$$
theta_m=n_mcdot 2 cdot pi=-2cdot 2 cdot pi=-12.6,,rm{rad}
$$

Result

4 of 4

$$
theta_h=-1.05,,rm{rad}
$$

$$
theta_m=-12.6,,rm{rad}
$$

Step 1

1 of 2

$Delta theta = (-2/12) (2pi) = -1.05 rad$

$$
Delta theta = (-2) (2pi) = -12.6 rad
$$

Result

2 of 2

a) $-1.05 rad$

b) $-12.6 rad$

Exercise 6

Step 1

1 of 2

Given:

$$
f=frac{1}{27.3},,rm{days^{-1}}
$$

$$
r=1.74cdot 10^6,,rm{m}
$$

a) Period in seconds can be calculated as:

$$
t=frac{1}{f}
$$

$$
t=27.3cdot 24cdot 60cdot 60
$$

$$
boxed{t=2358720,,rm{s}}
$$

b) Frequency can be calculated as:

$$
f=frac{1}{t}
$$

$$
boxed{f=frac{1}{2358720},,rm{s^{-1}}}
$$

c) Speed of the rock on the Moon’s equator is:

$$
v_c=2pi rf
$$

$$
v_c=frac{2cdot pi cdot 1.74cdot 10^6}{2358720}
$$

$$
boxed{v_c=4.64,,rm{m/s}}
$$

d) Speed of a person on the Earth’s surface:

$$
v_e=2pi r_e f_e
$$

$$
v_e=frac{2cdot pi cdot 6.4cdot 10^3}{24cdot 60cdot 60}
$$

$$
boxed{v_e=0.47,,rm{m/s}}
$$

Result

2 of 2

a) $t=2358720,,rm{s}$

b) $f=frac{1}{2358720},,rm{s^{-1}}$

c) $v_c=4.64,,rm{m/s}$

d) $v_e=0.47,,rm{m/s}$

Exercise 7

Step 1

1 of 4

The ball inside a computer mouse is rolled as mouse is moved a certain distance. We are given a diameter of a ball
$d=2,,rm{cm}=0.02,,rm{m}$
and a linear distance traveled
$s=12,,rm{cm}=0.12,,rm{m}$.
With this information we need to calculate angular displacement of the ball.

Step 2

2 of 4

In order to calculate angular displacement, we need to use an equation that connects linear and angular motion:
$$s=theta cdot frac{d}{2}$$
This equation says that linear distance traveled is proportional to the angular displacement multiplied with a factor of $dcdot pi$\
From the previous equation we can get angular displacement:
$$theta=frac{2cdot s}{d}$$

Step 3

3 of 4

Now when we put numbers we get:

$$theta=frac{2cdot 0.12}{0.02}$$
$$boxed{theta=12,,rm{rad}}$$

Result

4 of 4

$$theta=12,,rm{rad}$$

Exercise 8

Step 1

1 of 3

In this problem we need to analyse the minute hand of a clock. Two questions need to be answered; do all parts of the hand have the same angular displacement and to they all move same linear distance? \

Step 2

2 of 3

In order to solve this problem we must take a look at the way clock hand moves. \
Since it is made from a solid, not flexible material, we can assume that all of its parts maintain the same relative position to eachother.\
This means that each point on the clock hand is always set on the textbf{straight line and rotates as such}.
Which finally means that all parts will have textbf{the same angular displacement.}

Step 3

3 of 3

Since it is rotating as a straight line, we can take a look at the equation that connects angular displacement ($theta$), distance from the pivot point ($r$) and linear motion ($s$):
$$s=theta cdot r$$
From this we can conclude that by increasing radius, with constant angular displacement, we are textbf{increasing linear displacement.}

Exercise 9

Step 1

1 of 5

Drum of the washer spins at $omega=635,,rm{rev/min}$, it takes it $t=8,,rm{s}$ to slow to a stop. We need to calculate the angular acceleration of the drum that causes such decceleration.

Step 2

2 of 5

First thing we need to do is calculate angualr velocity in the SI units:

$$
omega=635,,rm{rev/min}cdot frac{2cdot pi ,,rm{rad/rev}}{60,,rm{s/min}}
$$

$$
omega=66.5,,rm{rad/s}
$$

Step 3

3 of 5

Next we know that angular acceleration can be calculated using an equation that connects angular acceleration, change in angular velocity and time:

$$
alpha=frac{omega_2-omega}{t}
$$

Where $omega_2=0,,rm{rad/s}$ because there is no angular velocity when the drum is stopped.

Step 4

4 of 5

Inserting values we get:

$$
alpha=frac{0-66.5}{8}
$$

Finally we get.

$$
boxed{alpha=-8.31,,rm{rad/s^2}}
$$

Result

5 of 5

$$
alpha=-8.31,,rm{rad/s^2}
$$

Exercise 10

Step 1

1 of 3

Given:

$$
begin{align*}
r_1&=2.7,,rm{cm}\
r_2&=5.5,,rm{cm}\
v&=1.4,,rm{m/s}
end{align*}
$$

a) Angular velocity of the disk for the start of the track can be calculated as:

$$
begin{align*}
omega_1 &=frac{v}{r_1}\
omega_1 &=frac{1.4}{2.7cdot 10^{-2}}
end{align*}
$$

$$
boxed{omega_1 =51.85,,rm{rad/s}}
$$

In rotations per minute:

$$
omega_1=51.85cdot frac{60}{2cdot pi}
$$

$$
boxed{omega_1=495,,rm{rpm}}
$$

b) Angular velocity of the disk for the end of the track can be calculated as:

$$
begin{align*}
omega_2 &=frac{v}{r_2}\
omega_2 &=frac{1.4}{5.5cdot 10^{-2}}
end{align*}
$$

$$
boxed{omega_2 =25.45,,rm{rad/s}}
$$

In rotations per minute:

$$
omega_2=25.45cdot frac{60}{2cdot pi}
$$

$$
boxed{omega_2=243,,rm{rpm}}
$$

Step 2

2 of 3

c) Given: $t=76,,rm{min}$

Which is $t=76cdot 60=4560,,rm{s}$

Now we can calculate acceleration:

$$
begin{align*}
alpha &=frac{omega_2-omega_1}{t}\
alpha &=frac{25.45-51.85}{4560}\
end{align*}
$$

$$
boxed{alpha=-5.8cdot 10^{-3},,rm{rad/s^2}}
$$

Result

3 of 3

a) $omega_1=51.85,,rm{rad/s}=495,,rm{rpm}$

b) $omega_2=25.45,,rm{rad/s}=243,,rm{rpm}$

c) $alpha=-5.8cdot 10^{-3},,rm{rad/s^2}$

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Page 200: Section Review

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