Question

$\sf{a \: particle \: moves \: on \: a \: circular \: path \: such} \\ \sf{ that \: angle \: rotated \: \theta \: depends \: on \: time} \\ \sf{t \: . \: according \: to \: equation \theta = 2 {t}^{2} \: find \: the \:} \\ \sf{acceleration \: on \: particle \: at \: t = 1s \: if \: at} \\ \sf{this \: moment \: its \: velocity \: is \: 8 \frac{m}{s}} . \\ \to \boxed{\bf{ \: answer = 8 \sqrt{17} }}$
​

Answer 1

Answer:

Given:

Angular displacement is given as :

$\theta = 2 {t}^{2}$

Velocity at t = 1 sec = 8 m/s

To find:

Total acceleration at t = 1 sec.

Concept:

We can very well understand that the object (undergoing circular motion) in this case will have both tangential acceleration as well as Centripetal acceleration.

Calculation:

$\bigstar \: \: \theta = 2 {t}^{2}$

$= > \omega = \dfrac{d \theta}{dt} = 4t$

$= > \alpha = \dfrac{d \omega}{dt} = 4 \: rad {s}^{ - 2}$

Now , at t = 1 second , we can say :

$\bigstar \: \: \omega = 4 \times (1)$

$= > \dfrac{v}{r} = 4$

$= > \dfrac{8}{r} = 4$

$= > r = 2 \: m$

Let the centripetal acceleration be a_{c}

$\bigstar \: \: a_{c} = \dfrac{ {v}^{2} }{r} = \dfrac{ {8}^{2} }{2} = 32 \: m {s}^{ - 2}$

Let tangential acceleration be a_{t}

$\bigstar \: \: a_{t} = (\alpha \times r) = 4 \times 2 = 8 \: m {s}^{ - 2}$

So total acceleration at t = 1 sec will be vector sum of centripetal acceleration and tangential acceleration:

$\bigstar \: \: a = \sqrt{ {(a_{t})}^{2} + {(a_{c})}^{2} }$

$= > a = \sqrt{ {(32)}^{2} + {(8)}^{2} }$

$= > a = \sqrt{1024 + 64}$

$= > a = \sqrt{1088}$

$= > a = 8 \sqrt{17} \: m {s}^{ - 2}$

Answer 2

SoluTion :

✴ Given :-

▪ A particle moves on a circular path such that angle rotated theta depends on time according to equation $\bf\theta=2t^2$

▪ Velocity of particle at t = 1sec is 8mps

✴ To Find :-

▪ Acceleration of particle at t = 1s

✴ Concept :-

✏ If the speed of the particle moving in a circle is not constant, the acceleration has both the radial and the tangential components.

✏ Formula of the radial and the tangential acceleration is given by

$\circ\:\bf{\red{a_r=\dfrac{v^2}{r}}}$

$\circ\:\bf{\blue{a_t=\alpha\times r}}$

✏ The magnitude of the acceleration is given by

$\circ\:\bf{\pink{a=\sqrt{{a_r}^2+{a_t}^2}}}$

✴ Calculation

$\mapsto\sf\:\theta=2t^2\\ \\ \mapsto\sf\:\omega=\dfrac{d\theta}{dt}=4t\\ \\ \mapsto\sf\:\alpha=\dfrac{d\omega}{dt}=4$

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▪ As per given data, Tangential velocity of particle at t = 1s is 8mps

▪ We know that, v = r × ω

$\leadsto\sf\omega=4t\\ \\ \leadsto\sf\dfrac{v}{r}=4(1)\\ \\ \leadsto\sf\dfrac{8}{r}=4\\ \\ \leadsto\bf\:radius \:of\:circular\:path=2\:m$

_________________________________

✒ Magnitude of acceleration at t = 1s

$\rightarrow\bf\:a=\sqrt{(\dfrac{v^2}{r})^2+(\alpha\times r)^2}\\ \\ \rightarrow\sf\:a=\sqrt{(\dfrac{64}{2})^2+(4\times 2)^2}\\ \\ \rightarrow\sf\:a=\sqrt{(32)^2+(8)^2}\\ \\ \rightarrow\sf\:a=\sqrt{1088}=\sqrt{64\times 17}\\ \\ \rightarrow\underline{\underline{\gray{\bf{a=8\sqrt{17}\:ms^{-2}}}}}$