A grindstone increases in angular speed from 4.00 rad / s to 12.00 rad / s in 4.00 s. Through what angle does it turn during that time interval if the angular acceleration is constant?

Question

A grindstone increases in angular speed from 4.00 rad / s to 12.00 rad / s in 4.00 s. Through what angle does it turn during that time interval if the angular acceleration is constant? (A) 8.00 rad (B) 12.0 rad (C) 16.0 rad (D) 32.0 rad

Tardigrade Admin · Accepted Answer

The angular displacement will be Δ θ =ω avg Δ t=((ωf+ωi/2)) Δ t =((12.00 rad / s +4.00 rad / s /2))(4.00 s )=32.0 rad

Q. A grindstone increases in angular speed from $4.00 r a d / s$ to $12.00 r a d / s$ in $4.00 s$ . Through what angle does it turn during that time interval if the angular acceleration is constant?

8.00 rad

12.0 rad

16.0 rad

32.0 rad

The angular displacement will be
$Δ θ = ω_{a vg} Δ t = (\frac{ω _{f} + ω _{i}}{2}) Δ t$
$= (\frac{12.00 r a d / s + 4.00 r a d / s}{2}) (4.00 s) = 32.0 r a d$

Q. A grindstone increases in angular speed from 4.00rad/s to 12.00rad/s in 4.00s. Through what angle does it turn during that time interval if the angular acceleration is constant?