Answer:
0.3086Nm
Explanation:
Given mass [tex]m=0.25kg[/tex], Radius [tex]r=0.25m[/tex]:
#Angular acceleration
[tex]\alpha=\frac{1700-0}{4.5}=377.78\ rpm/s=6.2963\ rot/s\\\\=6.2963\times 2\pi\\\\=39.5610rad/s\\[/tex]
Moment of inertia of the disk
[tex]I=0.5mr^2\\\\=0.5\times 0.25\times 0.25^2\\\\=0.0078 kg-m^2[/tex]
The torque required is therefore:
[tex]=I_\alpha=0.0078\times39.5610\\\\=0.3086\ Nm[/tex]
Hence, the torque required is 0.3086Nm
