Question

The second hand on a watch has a length of 4.50 mm and makes one revolution in 60.00 s. What is the speed of the end of the second hand as it moves in uniform circular motion

Answer 1

Answer:

$0.00015\pi m/s$ 0r 0.0004.71m/s

Explanation:

The speed v can be obtained through the following relationships;

$v=\omega l.................(1)$

where $\omega$ is its angular speed and $l$ is the length which is also equivalent to the amplitude of motion.

Also;

$\omega=\frac{2\pi}{T}..............(2)$

where T is the period of motion which is the time it takes for the second hand to make one complete revolution.

Substituting equation (2) into equation (1), we obtain the following;

$v=\frac{2\pi l}{T}..............(3)$

Given;

$l=4.5mm=0.0045m\\T=60.00s\\hence\\v=\frac{2\pi *0.0045}{60}..............(2)\\v=0.00015\pi m/s$

Taking $\pi=3.142$

$v=0.00015*3.142$

$v=0,0004.71m/s$

Answer 2

The speed of the end of the second hand as it moves in uniform circular motion is 4.7 × 10⁻⁴m/s.

Given the data in the question;

• Length of the second hand or radius; $r = 4.50mm = 0.0045 m$

• Time for one revolution; $t = 60.00s$

Speed of the end of the second hand; $v =\ ?$

First we determine the angular velocity of the second hand using the expression:

$w = \frac{2\pi }{t}$

Where ω is angular velocity and t is time taken.

We substitute in our given value

$w = \frac{2\pi }{60.00s}\\\\w = 0.1047rad/s$

Now, from Rotational kinematics:

$w = \frac{v}{r}$

Where ω is the angular velocity, r is the radius and v is the velocity

We substitute in our values and solve for "v"

$0.1047rad/s = \frac{v}{0.0045m} \\\\v = 0.1047rad/s * 0.0045m\\\\v = 0.00047 m/s\\\\v = 4.7*10^{-4}m/s$

Therefore, the speed of the end of the second hand as it moves in uniform circular motion is 4.7 × 10⁻⁴m/s.

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