Question

Rank the angular speed of the following objects, from highest to lowest:- A bowling ball of radius 12.3 cm rotating at 8.21 radians per second.- A tire of radius 0.321 m rotating at 75.8 rpm.- A 6.84 cm diameter top spinning at 375 degrees per second.- A square, with sides 0.458 m long, whose corners are moving with tangential speed 2.51 m/s as it rotates about its center.- A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration of 25.4 m/s2

Answer 1

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

Explanation:

To rank the angular speed (ω) of the objects, we need first calculate its value for every object:

A bowling ball of radius 12.3cm rotating at 8.21 radians per second:

ω = 8.21 rad/s

A tire of radius 0.321m rotating at 75.8 rpm:

`\omega = 75.8 (rev)/(min)\cdot (2\pi rad)/(1rev)\cdot (1min)/(60s) = 7.94rad/s`

A 6.84cm diameter top spinning at 375 degrees per second:

`\omega = 375 (^\circ)/(s)\cdot (2\pi rad)/(360^ \circ) = 6.54rad/s`

A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:

`\omega = (v)/(r) = (v)/((b)/(2)\sqrt 2) = (2.51 m/s)/((0.458 m)/(2) \sqrt 2) = 7.75rad/s`

A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:

`\omega = \sqrt (a)/(r) = \sqrt (25.4 m/s^(2))/(0.521m) = 6.98rad/s`

Now, the rank of the angular speed of the objects, from highest to lowest is:

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

I hope it helps you!