If the frequency of an object in uniform circ | vedclass

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

Question

(B) In uniform circular motion,the centripetal acceleration $a_c$ is given by the formula $a_c = \omega^2 r$,where $\omega$ is the angular velocity an

Vedclass · Accepted Answer

Four times

Vedclass · Answer

(B) In uniform circular motion,the centripetal acceleration $a_c$ is given by the formula $a_c = \omega^2 r$,where $\omega$ is the angular velocity and $r$ is the radius of the circular path.Since angular velocity $\omega = 2 \pi f$,where $f$ is the frequency,we can substitute this into the acceleration formula:$a_c = (2 \pi f)^2 r = 4 \pi^2 f^2 r$.From this expression,we can see that the acceleration is directly proportional to the square of the frequency: $a_c \propto f^2$.If the frequency $f$ is doubled $(f' = 2f)$,the new acceleration $a_c'$ becomes:$a_c' \propto (2f)^2 = 4f^2$.Therefore,the acceleration becomes $4$ times the original value.

If the frequency of an object in uniform circular motion is doubled,its acceleration becomes

Similar Questions

The relative angular speed of the hour hand and the second hand of a clock is (in $rad/s$):

$A$ wheel rotates with a constant acceleration of $2.0 \ rad/s^2$. If the wheel starts from rest,the number of revolutions it makes in the first ten seconds is:

Obtain the equation $\omega = \omega_{0} + \alpha t$ from first principles.

The relative angular speed of the hour hand and minute hand of a clock is (in $rad/s$):

$A$ wheel has a speed of $1200$ revolutions per minute and is made to slow down at a rate of $4 \ rad/s^{2}$. The number of revolutions it makes before coming to rest is

Vedclass Test Series

Exam Paper Generator

Online Exam Module