Question

The minute hand of a large clock is 0.50 m long. (A) Calculate its linear speed at its tip in meters per second. (0.000873 m/s)
(B) Calculate the centripetal acceleration of the tip of the hand. (1.52 x 10-6 m/s2)

Answer 1

The rotational speed of the minute hand is 360 degrees in one hour or:
ω = 2π rad/3600 sec
or 
ω = 0.001745 rad/sec

To convert this to linear speed, we use the formula:
v = rω

Substituting the given values:
v = 0.5 (0.001745) = 0.000873 m/s

For the centripetal acceleration:
a = v^2/r
a = (0.000873)^2 / 0.5
a = 1.52 x 10^-6 m/s^2

Answer 2

This question involves the concepts of linear speed and centripetal acceleration.

(a) The linear speed of the minute hand is "0.000873 m/s".

(b) The centripetal acceleration of the tip of the hand is "1.52 x 10⁻⁶ m/s²".

(a) LINEAR SPEED

The linear speed is given by the following formula:

$v=r\omega$

where,

• v = linear speed = ?
• r = radius = length of minute hand = 0.5 m
• ω = angular speed of minute hand = $\frac{2\pi rad}{t}=\frac{2\pi\ rad}{3600\ s}=0.00174\ rad/s$

Therefore,

v = (0.5\ m)(0.00174 rad/s)

v = 0.000873 m/s

(b) CENTRIPETAL ACCELERATION

Centripetal acceleration is given by the following formula:

$a=\frac{v^2}{r}=\frac{(0.000873\ m/s)^2}{0.5\ m}$

a = 1.52 x 10⁻⁶ m/s²

Learn more about centripetal acceleration here:

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