A wheel starts from rest is uniformly accelerated at 2 rad/s^2 for 20s. It is allowed to rotate uniformly for the next 10s and is finally brought to rest in next 20s. Find the total angle rotated by the wheel in radian.

  • 105
Applying rotational analogue of s=ut+0.5 at^2, we get angular displacement of the accelerated period as (0+0.5*2*20)=400 radians and angular velocity attained is angular acc*time = 40 rads^-1. In 10s of uniform motion, ang displ is 40*10 = 400 rad. Decelerated motion ang displ is same as first case, 400 radians. Add the three 2gether,total disp= 1200 rad.
  • 26
corrrect answer is 1200 rad...
 
  • -17

ω = 0 + 2(20) = 40               (according to rotational equation ω = ω0 + αt)

and θ = 0 + 1/2(2)(20)2 = 400 radian

in middle uniform part angle rotated = 40*10 = 400 radian

and since acceleration in first part and retardation in last part are equal so angle rotated in last part = 400 radian

total = 1200 radian

  • 23
1200rad.
  • -18
for first part:->
As we know : s=ut+1/2 at^2
for rotational analogue :  θ = 0 + 1/2(2)(20)2 = 400 radian;-----(1)

for second part:->
ω = 0 + 2(20) = 40 (according to rotational equation ω = ω0 + αt)--- angular velocity attained during first part of its motion
we got ω = 40 
As angular velocity is uniform 
we use θ = ωt - 40 * 10 = 400 rad ;------(2)

for third part:->
As you have seen in first case it goes from 0 to 20 sec with 2 rad/s^2
it will decelerate with 2 rad/s^2 for 20 sec
so θ = 400 rad--------(3)
add all three you wil get 1200 radian
  • 19
400+400+400=1200 rad
  • -6
12000
  • -7
? = 0 + 2(20) = 40 ? ? ? ? ? ? ? (according to rotational equation?? =??0?+??t)

and?? = 0 + 1/2(2)(20)2?=?400 radian

in middle uniform part angle rotated = 40*10 = 400 radian

and since acceleration in first part and retardation in last part are equal so angle rotated in last part = 400 radian

total = 1200 radian

l hope, it helps.
  • 5
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