A hollow, thin-walled sphere of mass 10.0 kg and diameter 51.5 cm is rotating ab

Question

A hollow, thin-walled sphere of mass 10.0 kg and diameter 51.5 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by

(t) = At2 + Bt4,

where A has numerical value 1.41 and B has numerical value 1.09.

(a) What are the units of the constants A and B?

(b) At the time 3.00 s, find the following.

(i) the angular momentum of the sphere
kg · m2/s

(ii) the net torque on the sphere
N · m

Explanation / Answer

a)
(t) = At^2 + Bt^4
unit of is rad
so,
unit of A = rad/s^2
unit of B = rad/s^4

b)
(t) = At^2 + Bt^4

(t) = 1.41*t^2 + 1.09*t^4
w(t) = d/dt ((t))
= 1.41*2*t + 1.09*4*t^3
=2.82*t + 4.36*t^3

put t = 3 s
w = 2.82*3 + 4.36*3^3
= 126.18 rad/s

r = 51.5/2 cm = 25.75 cm = 0.2575 m

I = (2/5)*m*r^2
= (2/5)*10*(0.2575)^2
= 0.265 Kgm^2

angular momentum = I*w
= 0.265*126.18
=33.5 Kgm^2/s

c)
a(t) = d/dt(w(t))
= d/dt (2.82*t + 4.36*t^3)
= 2.82 + 4.36*3*t^2
= 2.82 + 13.08*t^2
at t = 3 s
a = 2.82 + 13.08*3^2
=120.54 rad/s^2

use:
T = I*a
= 0.265*120.54
=31.9 Nm
Answer: 31.9 Nm

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