(A) Given, $\theta = a t^2$.The angular velocity is $\omega = \frac{d\theta}{dt} = 2at$.The tangential velocity is $v = \omega r = 2atr$. Thus, $\frac

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

$\frac{v}{t} \sqrt{1+4 a^2 t^4}$

(A) Given, $\theta = a t^2$.The angular velocity is $\omega = \frac{d\theta}{dt} = 2at$.The tangential velocity is $v = \omega r = 2atr$. Thus, $\frac{v}{t} = 2ar$.The tangential acceleration is $a_t = \frac{dv}{dt} = 2ar$.The radial (centripetal) acceleration is $a_n = \frac{v^2}{r} = \frac{(2atr)^2}{r} = 4a^2t^2r$.The total acceleration is $a_{\text{total}} = \sqrt{a_t^2 + a_n^2} = \sqrt{(2ar)^2 + (4a^2t^2r)^2}$.$a_{\text{total}} = \sqrt{4a^2r^2 + 16a^4t^4r^2} = 2ar \sqrt{1 + 4a^2t^4}$.Since $2ar = \frac{v}{t}$, we have $a_{\text{total}} = \frac{v}{t} \sqrt{1 + 4a^2t^4}$.

Class 11 Physics 3-2.Motion in Plane Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

Medium

Consider a wheel rotating around a fixed axis. If the rotation angle $\theta$ varies with time as $\theta=a t^2$, then the total acceleration of a point $A$ on the rim of the wheel is ($v$ being the tangential velocity).

TS EAMCET 2018 All TS EAMCET

A
$\frac{v}{t} \sqrt{1+4 a^2 t^4}$
B
$\frac{v}{t}$
C
$\frac{v}{t}\left(1+4 a^2 t^4\right)$
D
$\sqrt{\left(1+4 a^2 t^4\right)}$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

3-2.Motion in Plane

Subtopic

Non-uniform Circular Motion (Centrifugal/Centripetal and tangential Accelaration)

Previous Year

TS EAMCET 2018 Paper All TS EAMCET years →

$A$ particle moves along a circle of radius $3 \, m$ with its displacement given by $S = \frac{t^2}{2} + \frac{t^3}{3}$. The total acceleration at $t = 2 \, s$ is ....... $m/s^2$.

Difficult

View Solution

$A$ particle moves in a circular path of radius $r$ with speed $v.$ It then increases its speed to $2v$ while travelling along the same circular path. The centripetal acceleration of the particle has changed by a factor of

Easy

View Solution

In a hypothetical ring-shaped satellite rotating in space, artificial gravity can be achieved using centripetal force. If the satellite has a radius of $10 \,m$, then to achieve a centripetal acceleration at a point on the circumference of $10 \,ms^{-2}$, its angular speed is:

EasyAP EAMCET 2022

View Solution

$A$ particle at rest starts moving with a constant angular acceleration of $4 \ rad/s^2$ in a circular path. At what time will the magnitudes of its tangential acceleration and centrifugal acceleration be equal (in $s$)?

EasyMHT CET 2022

View Solution

$A$ particle is moving in a circle of radius $50 \ cm$ in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at $t=0$ is $4 \ m/s$,the time taken to complete the first revolution will be $\frac{1}{\alpha}[1-e^{-2 \pi}] \ s$,where $\alpha=$ . . . . . . .

DifficultJEE MAIN 2024

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Consider a wheel rotating around a fixed axis. If the rotation angle $\theta$ varies with time as $\theta=a t^2$, then the total acceleration of a point $A$ on the rim of the wheel is ($v$ being the tangential velocity).

Similar Questions

$A$ particle moves along a circle of radius $3 \, m$ with its displacement given by $S = \frac{t^2}{2} + \frac{t^3}{3}$. The total acceleration at $t = 2 \, s$ is ....... $m/s^2$.

$A$ particle moves in a circular path of radius $r$ with speed $v.$ It then increases its speed to $2v$ while travelling along the same circular path. The centripetal acceleration of the particle has changed by a factor of

In a hypothetical ring-shaped satellite rotating in space, artificial gravity can be achieved using centripetal force. If the satellite has a radius of $10 \,m$, then to achieve a centripetal acceleration at a point on the circumference of $10 \,ms^{-2}$, its angular speed is:

$A$ particle at rest starts moving with a constant angular acceleration of $4 \ rad/s^2$ in a circular path. At what time will the magnitudes of its tangential acceleration and centrifugal acceleration be equal (in $s$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module