Question
A rigid triangular body, PQR, with sides of equal length of 1 unit moves on a flat plane. At the instant shown, edge QR is parallel to the x-axis, and the body moves such that velocities of points P and R are VP and VR, in the x and y directions, respectively. The magnitude of the angular velocity of the body is
Options :
VR / √3
VP / √3
2VR
2VP
Answer :
2VR
Solution :
The given body is a rigid body; Hence the given body will not have any deformation, and the possible motion is translation and angular motion. we know that the angular motion of the object is always defined as the motion about the point. Hence, first we have to find a point parallel to both velocity VP and VR.
∴ Angular velocity, ω = Velocity / Perpendicular distance
ω = VP / PS = VR /SR
PS = (√3 / 2) × Side of triangle = (√3 /2) × 1 = √3/ 2
SR = 1 /2 × Side of triangle = 1 /2 × 1 = 1/ 2
ω = 2VP /√3 = 2VR/ 1
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