In the question number 76, the force acting on the particle is

Question

A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 m $s^{- 1}$ and the speed is increasing at a rate of 2 m $s^{- 2}$ . The magnitude of net acceleration at this instant is, the force acting on the particle is

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(2)

Given: $\vec{r} = A \cos ω t \hat{i} + B \sin ω t \hat{j}$

Velocity, $\vec{v} = \frac{d \vec{r}}{d t} = \frac{d}{d t} (A \cos ω t \hat{i} + B \sin ω t \hat{j})$

$= - A ω \sin ω t \hat{i} + B ω \cos ω t \hat{j}$

Acceleration, $\vec{a} = \frac{d \vec{v}}{d t} = - A ω^{2} \cos ω t \hat{i} - B ω^{2} \sin ω t \hat{j}$

$\vec{a} = - ω^{2} [A \cos ω t \hat{i} + B \sin ω t \hat{j}] = - ω^{2} \vec{r}$

The force acting on the particle is

$\vec{F} = m \vec{a} = m (- ω^{2} \vec{r}) = - m ω^{2} \vec{r}$

Solution

Q.
A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 m $s^{- 1}$ and the speed is increasing at a rate of 2 m $s^{- 2}$ . The magnitude of net acceleration at this instant is, the force acting on the particle is

1 $m ω^{2} \vec{r}$

2 $- m ω^{2} \vec{r}$

3 $2 m ω^{2} \vec{r}$

4 $- 2 m ω^{2} \vec{r}$

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Solution

Q. A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 m s-1 and the speed is increasing at a rate of 2 m s-2. The magnitude of net acceleration at this instant is, the force acting on the particle is

1 mω2r→

2 -mω2r→

3 2mω2r→

4 -2mω2r→