In a jungle safari, a cheetah (leopard) was observed running along a straight track at a constant velocity of 125m/s. A fixed TV camera is recording the event as shown in figure. In order to keep the cheetah in view, the position shown, the angular velocity of camera should be
(a)10rad/s (b) 4.5rad/s
(c)5rad/s (d) 8rad/s
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Hades09:
what is the answer? I'll provide the solution
(d) 8rad/s
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Refer to the diagram above.
Cheetah is moving with a speed of 125m/s.
DH = 10m. We have to find ω (angular velocity of the TV camera).
Using Trigonometry,
DH = EH.cos(37°)
10 = EH(0.8)
EH = 12.5
Velocity of Cheetah alone the line EG:
(Note: Note that the velocity of Cheetah along EG is the tangential velocity of a particle moving with 125m/s in a circular orbit centered at H with radius EH.)
125.cos(37°)=Velocity along EG
Velocity along EG = 100m/s
Now,
v = (r) x (ω)
ω = v/r
ω = (100)/(12.5)
ω = 8rad/sec
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Cheetah is moving with a speed of 125m/s.
DH = 10m. We have to find ω (angular velocity of the TV camera).
Using Trigonometry,
DH = EH.cos(37°)
10 = EH(0.8)
EH = 12.5
Velocity of Cheetah alone the line EG:
(Note: Note that the velocity of Cheetah along EG is the tangential velocity of a particle moving with 125m/s in a circular orbit centered at H with radius EH.)
125.cos(37°)=Velocity along EG
Velocity along EG = 100m/s
Now,
v = (r) x (ω)
ω = v/r
ω = (100)/(12.5)
ω = 8rad/sec
If you like this answer, rate it as the Branliest answer. Thanks.
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refer to attachment....!!!
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