A particle is performing U.C.M. along the cir | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given: Diameter of the circle,$d = 50 \ cm = 0.5 \ m$.Radius of the circle,$r = d/2 = 0.25 \ m$.Frequency of the particle,$f = 2 \ Hz$.In uniform

Vedclass · Accepted Answer

$4 \pi^2$

Vedclass · Answer

(D) Given: Diameter of the circle,$d = 50 \ cm = 0.5 \ m$.Radius of the circle,$r = d/2 = 0.25 \ m$.Frequency of the particle,$f = 2 \ Hz$.In uniform circular motion $(U.C.M.)$,the acceleration is the centripetal acceleration,given by $a = \omega^2 r$.Angular frequency,$\omega = 2 \pi f = 2 \pi (2) = 4 \pi \ rad/s$.Substituting the values into the formula:$a = (4 \pi)^2 	imes 0.25$$a = 16 \pi^2 	imes 0.25$$a = 4 \pi^2 \ m/s^2$.

$A$ particle is performing $U.C.M.$ along the circumference of a circle of diameter $50 \ cm$ with frequency $2 \ Hz$. The acceleration of the particle in $m/s^2$ is

Similar Questions

Decomposers are

The moment of inertia of a meter scale of mass $0.6\, kg$ about an axis perpendicular to the scale and located at the $20\, cm$ position on the scale in $kg\, m^2$ is (Breadth of the scale is negligible).

Which of the following is a homosporous pteridophyte?

The radius of the circle $r = \sqrt{3} \sin \theta + \cos \theta$ is

Identify the wrong statement in the following.

Vedclass Test Series

Exam Paper Generator

Online Exam Module