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TS EAMCET · Physics · Motion In Two Dimensions

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)

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

Answer & Solution

Correct Answer

(A) \(\frac{v}{t} \sqrt{1+4 a^2 t^4}\)

Step-by-step Solution

Detailed explanation

Given, \(\theta=a t^2\) So, angular velocity, \(\omega=\frac{d \theta}{d t}=2 a t\) Linear tangential velocity, \[ v=\omega r \quad=2 a t r \quad \frac{v}{t}=2 a r \] So, tangential linear acceleration of particle is \[ a_t=\frac{d v}{d t}=2 a r \] and angular acceleration of…
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