Answer:
a)- 1.799 rad/sec²
b)- 17.6 x 10ˉ³Nm
Explanation:
ω₀ = 720 rev/min x (1 min/60 sec) x (2π rad / 1 rev) = 24π rad/s
a) Assuming a constant angular acceleration, the formula will be
α = (ωf -ω₀) / t
As final state of the grindstone is at rest, so ωf =0
⇒ α = (0-24π) / 41.9 = - 1.799 rad/sec²
b)Moment of inertia I for a disk about its central axis
I = ½mr²
where m=2kg and radius 'r'= 0.099m
I = ½(2)(0.099²)
I = 9.8 x 10ˉ³ kgm²
Next is to determine the frictional torque exerted on the grindstone, that caused it to stop, applying the rotational equivalent of the Newton's 2nd law:
τ = I α =>(9.8 x 10ˉ³)(- 1.799)
τ = - 17.6 x 10ˉ³Nm
(The negative sign indicates that the frictional torque opposes to the rotation of the grindstone).
Figure B. The second option.
The formula for mass you can find from the density equation. Multiply the volume by both sides of the equation.
density = mass/volume
volume*density = mass
And there you are :)
Part (a): Specific volume
Specific volume, v = V/m, V = Volume of the tank, m = mass of CO in the tank
Therefore,
v = 1/4 = 0.25 m^3/kg
Part (b): Energy transferred in kJ
Work done, W = PΔt, where P = power = 14 W = 14 J/s, t = time = 1 hour = 60*60 = 3600 seconds
Therefore,
W = 14*3600 = 50400 J = 50.4 kJ
Part (c): Energy transferred by heat
ΔU = Q + W
Then,
Heat transferred by heat, Q = ΔU - W
But, ΔU = mΔu = 4 kg*10 kJ/kg = 40 kJ
Therefore,
Q = 40 - 50.4 = -10.4 kJ (negative sign indicates that heat is removed from the CO).
