Answer
given,
initial speed of merry-go-round = 0 rad/s
final speed of merry-go-round = 1.5 rad/s
time = 7 s
Radius of the disk = 6 m
Mass of the merry-go-round = 25000 Kg
Moment of inertia of the disk
[tex]I = \dfrac{1}{2}MR^2[/tex]
[tex]I = \dfrac{1}{2}\times 25000\times 6^2[/tex]
  I = 450000 kg.m²
angular acceleration
[tex]\alpha = \dfrac{\omega_f-\omega_0}{t}[/tex]
[tex]\alpha = \dfrac{1.5-0}{7}[/tex]
[tex]\alpha =0.214\ rad/s^2[/tex]
we know,
[tex]\tau= I \alpha[/tex]
[tex]\tau= 450000\times 0.214[/tex]
[tex]\tau=96300\ N.m[/tex]
The required value of frictional torque produced by the bearings is 96300 N-m.
Given data:
The final angular speed of merry-go round is, [tex]\omega_{2}=1.5 \;\rm rad/s[/tex].
The time interval is, t = 7.0 s.
The radius of merry-go round is, r = 6.0 m.
The mass of merry-go round is, m = 25,000 kg.
The frictional torque of an object undergoing rotational motion is the product of moment of inertia and angular acceleration. So,
[tex]T_{f} = I \times \alpha[/tex] ....................................................(1)
Here, I is the moment of inertia of merry-go round and its value is,
[tex]I =\dfrac{mr^{2}}{2}\\\\I =\dfrac{25000 \times (6.0)^{2}}{2}\\\\I =450000 \;\rm kg.m^{2}[/tex]
And angular acceleration is,
[tex]\alpha = \dfrac{\omega_{2}-\omega_{1}}{t}\\\\\alpha = \dfrac{1.5-0}{7}\\\\\alpha = 0.214 \;\rm rad/s^{2}[/tex]
Then the frictional torque is calculated from equation (1) as,
[tex]T_{f} = 450000 \times 0.214\\\\T_{f}=96300 \;\rm N.m[/tex]
Thus, we can conclude that the required value of frictional torque produced by the bearings is 96300 N-m.
Learn more about the frictional torque here:
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