A grinding wheel, initially at rest, is ro-tated with constant angular acceleration of2.52 rad/s2for 7.03 s. The wheel is thenbrought to rest with uniform deceleration in11.9 rev. Find the angular acceleration required tobring the wheel to rest. Note that an in-crease in angular velocity is consistent with apositive angular acceleration. Answer in units of rad/s2.

Starting from rest, the wheel accelerates to an angular velocity of

ω = (2.52 rad/s²) (7.03 s) ≈ 17.7 rad/s

then undergoes a new acceleration α until it comes to a rest. It does so in 11.9 revolutions, or with an angular displacement of (11.9 rev) • (2π rad/rev) = 23.8π rad. So α satisfies

0² - ω² = 2 α (23.8π rad)

α = - ω² / (47.6π rad)

α ≈ -2.10 rad/s²

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

which piece of furniture will have the most inertia and give the furniture movers the most difficulty in moving?  

answer

a 75kg dining table