A dentist\'s drill starts from rest. After 3.20 s of constant angular accelerati
ID: 1287395 • Letter: A
Question
A dentist's drill starts from rest. After 3.20 s of constant angular acceleration it turns at a rate of 2.2 Times 104 rev/min. Find the drill's angular acceleration. rad/s2 Determine the angle (in radians) through which the drill rotates during this period, rad 1+1 -12 points SerCP9 7.P.010.soln.The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.0 rev/s in 10.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 14.0 s. Through how many revolutions does the tub turn during this 24-s interval? Assume constant angular acceleration while it is starting and stopping.+ -12 points SerCP9 7.P.017.soln.What is the tangential acceleration of a bug on the rim of a 12.0-in.-diameter disk if the disk accelerates uniformly from rest to an angular speed of 77.0 rev/min in 5.00 s? m/s2 When the disk is at its final speed, what is the tangential velocity of the bug? m/s One second after the bug starts from rest, what is its tangential acceleration? One second after the bug starts from rest, what is its centripetal acceleration?(One second after the bug starts from rest, what is its total acceleration? from the radically inward directionExplanation / Answer
1a)Angular acceleration=w/t=2.2*10^4*2pi/(60*3.2)=719.95rad/s^2
b)theta=0.5*alpha*t^2=0.5*719.95*3.2^2=3686.14 rad
2a)alpha1=w/t=3*2pi/10=1.885rad/s^2
alpha2=3*2pi/14=1.3464rad/s^2
theta=0.5alpha1*t1^2+0.5*alpha2*t2^2=0.5*1.885*10^2+0.5*1.3464*14^2=226.20rad
3a)a=alpha*r=wr/t=(77*2pi/60)*(12*0.0254)/5=0.4915m/s^2
b)v=wr=(77*2pi/60)*(12*0.0254)=2.46m/s
c)a=0.4915m/s^2 (constant acceleration)
d)ac=w^2r=(alpha*t)^2*r=(77*2pi*1/(60*5))^2*(12*0.0254)=0.7927m/s^2
Total a=sqrt(0.7927^2+0.4915^2)=0.9327 m/s^2
Direction=Tan^-1(0.4915/0.7927)=31.8 degrees from the radially inward direction
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