Answer: The torque required is 0.0471 N m.
Explanation:
Mass of the disc = 200 g = 0.2 kg (1 kg =1000 g)
Radius of the disc =[tex]\frac{Diameter}{2}[/tex]= 10 cm = 0.1 m(1 m = 100 cm)
Angular acceleration = [tex]\alpha [/tex]
[tex]\alpha =\frac{2\pi\times 1800}{60 \times 4 sec}=15\pi rad/s^2[/tex]
Moment of inertia = [tex]\frac{1}{2}\times mass\times (radius)^2=\frac{1}{2}\times0.2 kg\times (0.1 m)^2=0.001 kg m^2[/tex]
[tex]Torque=\alpha \times I=0.001 kg m^2\times 15\times 3.14 rad/s^2r=0.0471 N m[/tex]
The torque required is 0.0471 N m.
The value of the torque in the question is given as 0.0471N/m
The mass = 200 grams
Convert to kg = 0.2 kg
Radius = diameter/2
= 20/2 = 10cm
convert to meters = 0.1m
1800/4 = 450
450/60sec
= 7.5rot/sec²
7.5*2π
= 47.14
I = 1/2mr²
= 1/2*0.2*0.1²
= 0.0001kg
= 0.001 * 47.14
= 0.04714N.M
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