Answer:
The mass of the solid cylinder is [tex]m = 1612.5 \ kg[/tex]
Explanation:
From the question we are told that
The radius of the grinding wheel is [tex]R = 0.330 \ m[/tex]
The tangential force is [tex]F_t = 250 \ N[/tex]
The angular acceleration is [tex]\alpha = 0.940 \ rad/s^2[/tex]
The torque experienced by the wheel is mathematically represented as
[tex]\tau = I * \alpha[/tex]
Where I is the moment of inertia
The torque experienced by the wheel can also be mathematically represented as
[tex]\tau = F_t * r[/tex]
substituting values
[tex]\tau = 250 * 0.330[/tex]
[tex]\tau = 82.5 \ N\cdot m[/tex]
So
[tex]82.5 = I * \alpha[/tex]
[tex]82.5 = I * 0.940[/tex]
So
[tex]I = 87.8 \ kg \cdot m^2[/tex]
This moment of inertia can be mathematically evaluated as
[tex]I = \frac{1}{2} * m* r^2[/tex]
substituting values
[tex]87.8 = \frac{1}{2} * m* (0.330)^2[/tex]
=> [tex]m = 1612.5 \ kg[/tex]
