Question

A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.

A-What is the rotational speed of the ladybug? (in rev/sec)
B-What is the frequency of the ladybug's revolutions? (in Hz)
C-What is the tangential speed of the ladybug? (in cm/sec)
D-After 20 seconds. how far has the ladybug traveled? (in cm)

Answer 1

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

$\omega = 25 rpm$

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

$\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s$

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

$\omega = 2 \pi f$ (1)

where

$\omega$ is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

$\omega=0.42 rev/s$

Keeping in mind that $1 rev = 2\pi rad$, the angular speed can be rewritten as

$\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42$

And re-arranginf eq.(1), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz$

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

$v=\omega r$

where

$\omega$ is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

$\omega = 2\pi \cdot 0.42 rad/s$ is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

$v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s$

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

$d=vt$

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

$d=(26.4)(20)=528 cm$

Learn more about circular motion:

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#LearnwithBrainly

Answer 2

Answer:

A $w= 0.417 \ rev/sec$

B f = 0.417 Hz

C vt = 26.2 cm/s

D x = 524 cm

Explanation:

Circular Motion

This is one case where the object rotates around a fixed point called center at the same rate, i.e. forms similar angles at similar times. When this happens, the angular speed is constant, or

w=constant

The units of w are rad/sec, rev/min, rev/sec and other similar ones.  

A. We are given the rotational speed in rev/min (rpm), to change it to rev/sec, we have to convert the minute to seconds:

B. The frequency of the ladybug's revolutions is given by

We must express the angular speed in rad/sec, so

$w=0.417\ rev/sec \times 2\pi$

$w=2.62\ rad/sec$

$\displaystyle f=\frac{2.62}{2\pi}$

$f=0.417 \ Hz$

C. The tangential speed of the ladybug is calculated with the formula

where r is the distance from the object (ladybug) to the center, r=10 cm, thus

$v_t=26.2\ m/s$

D. The tangential speed will make the ladybug traveled a distance which can be computed with the formula

$x=v_t.t$

$x=26.2\times 20$

$x=524\ cm$