A grinding wheel is in the form of a uniform solid disk of radius 6.99 cm and mass 2.08 kg. It starts from rest and accelerates uniformly under the action of the constant torque of 0.591 N · m that the motor exerts on the wheel.(a) How long does the wheel take to reach its final operating speed of 1 230 rev/min? s(b) Through how many revolutions does it turn while accelerating? revNeed Help?

A) 1.11 secs

B) 11.41 revolutions

Explanation:

A) We are given final operating speed; ω_f = 1230rev/min = 1230 x 2π/60 rad/s = 128.81 rad/s

Torque; τ = 0.591 N

Mass;m = 2.08 kg

radius; r = 6.99 cm = 0.0699m

Let's now find the acceleration from;

torque = inertia x acceleration

Inertia = (1/2)mr²

Thus;

τ = (½)(mr²)a

So,

0.591 = ½(2.08 x 0.0699²)a

0.591 = 0.00508a

a = 116.34rad/s²

using the equation of rotational motion;

ω_f = ω_i + at

ω_i is initial angular velocity and it's zero.

Thus;

128.81 = 0 + 116.34t

128.81 = 116.34t

t = 1.11secs

B) the number of revolutions will be calculated from;

θ = (ω_i)t + ½at²

θ = 0(1.81) + ½(116.34)(1.11)²

θ = ½(116.34)(1.11)²

θ = 71.671257 rad

Let's convert it to revs

θ = 71.671257 ÷ 2π

θ = 11.41 revolutions

i would think it's food chain

0.11 m/s and 0.28 m/s should be correct