1) [tex]-9.3rad/s^2[/tex]
2) 3.0 s
Explanation:
1)
The angular acceleration of a rigid body in rotation can be found by using the equivalent of Newton's second law for rotational motions:
[tex]\tau = I\alpha[/tex] (1)
where
[tex]\tau[/tex] is the torque on the object
I is the moment of inertia of the object
[tex]\alpha[/tex] is the angular acceleration
Here we have:
[tex]I=\frac{2}{5}MR^2[/tex] is the moment of inertia of a solid sphere about its central axis, where
M = 240 g = 0.240 kg is the mass of the sphere
R = 4.50 cm /2= 2.25 cm = 0.0225 m is the radius of the sphere
[tex]\tau = F r[/tex] is the torque exerted by the frictional force, where
[tex]F=-0.0200 N[/tex] is the force of friction (negative because the direction is opposite to the motion)
r = R = 0.0225 m is the distance of the point of application of the force from the centre
Substituting into eq(1) we find
[tex]FR=\frac{2}{5}MR^2 \alpha[/tex]
And solving for [tex]\alpha[/tex], we find the angular acceleration:
[tex]\alpha = \frac{5F}{2MR}=\frac{5(-0.0200)}{2(0.240)(0.0225)}=-9.3rad/s^2[/tex]
2)
Here we know that the motion of the sphere is an angular accelerated motion.
Therefore, we can use the equivalent of suvat equations for rotational motion:
[tex]\omega_f = \omega_i +\alpha t[/tex]
where
[tex]\omega_f[/tex] is the final angular velocity
[tex]\omega_i[/tex] is the initial angular velocity
[tex]\alpha[/tex] is the angular acceleration
t is the time
In this problem, we know that
[tex]\omega_f-\omega_i = -28.0 rad/s[/tex], since we are told that the rotational speed decreases by 28.0 rad/s
[tex]\alpha=-9.3 rad/s^2[/tex] is the angular acceleration of the sphere
Solving for t, we find how long it takes for the sphere to decelerate by 28.0 rad/s:
[tex]t=\frac{\omega_f - \omega_i}{\alpha}=\frac{-28.0}{-9.3}=3.0 s[/tex]
