Answer:
Time taken by disk to stop spinning will be 4.60 sec
Explanation:
It is given mass m = 7.50 kg
Diameter of the disk d = 88 cm = 0.88 m
So radius of the disk [tex]r=\frac{d}{2}=\frac{0.88}{2}=0.44m[/tex]
Angular speed of the disk [tex]\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 280}{60}=29.30rad/sec[/tex]
Force is given F = 35 N
Coefficient of friction [tex]\mu =0.3[/tex]
Friction force will be equal to [tex]f=\mu F=0.3\times 35=10.5N[/tex]
So torque will be equal to [tex]\tau =Fr=10.5\times 0.44=4.62Nm[/tex]
Torque is also equal to [tex]\tau =I\alpha[/tex]
Moment of inertia of the disk [tex]I=\frac{1}{2}Mr^2=\frac{1}{2}\times 7.5\times 0.44^2=0.726kgm^2[/tex]
So [tex]4.62=0.726\times \alpha[/tex]
[tex]\alpha =6.964rad/sec^2[/tex]
So time taken will be equal to [tex]t=\frac{\omega -0}{\alpha }=\frac{29.30-0}{6.964}=4.60sec[/tex]
