Question

4. A particle is moving with constant speed 10 m/s in
x-y plane as shown in the figure. The magnitude of
its angular velocity about origin at this instant is.
(x and y are in m)
V:

(1) 1.4 rad/s
(2) 1.2 rad/s
(3) 1.6 rad/s
(4) 2.2 rad/s​

Answer 1

(3)1.6 rad/sec

Angular velocity is given as :

Omega =v/r

As velocity is Tangential in circular motion.

Omega =v*sin(theta)/root(a²+b²)

Sin ( theta)= b/root(a²+b²)

So, Omega =v*b/a²+b²

Omega= 10*4/3²+4² =40/25

= 1.6 rad/sec

Answer 2

the angular velocity with reference to the origin in the following case is equal to

(3)1.6 rad/sec

Explanation:

• angular velocity is the ratio between the tangential velocity and the perpendicular distance

• here the perpendicular distance is the hypotenuse of the triangle made by x and y axis with the origin. This can be found out by the Pythagoras theorem.

• If a=3 and b= 4, the perpendicular distance r=(a²+b²)= (3²+4² ) = 25

• The tangential velocity v= vxb=10x4= 40m / sec.

• The angular momentum,ω= 40/25 rad/ sec.

• This is equal to 1.6rad/ sec.

To know more about angular velocity

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