Name: Inertial Frames of Reference
Uploaded: 2019-07-19
Duration: 7 min 10 s
Description: Imagine if you are on a bus or a train, can you say you are in an inertial frame of reference? If yes, how can you predict it? Let us explore more in this vi...

Question

A point traversed half a circle of radius

R = 160 cm

during time interval

τ = 10.0 s

. Calculate the following quantities averaged over that time.

(a)

the mean velocity

⟨ v ⟩

(b)

the modulus of the mean velocity vector

| ⟨ \vec{v} ⟩ |

(c)

the modulus of the mean vector of the total acceleration

| ⟨ \vec{w} ⟩ |

if the point moved with constant tangent acceleration.

Accepted Answer

Given,

radius of the circle, $R = 160 cm$ ,

duration of motion, $τ = 10.0 s$ .

$(a)$ Mean velocity or average speed is given by,
$<v> = \frac{Total distance travelled}{Time duration}$ $= \frac{S}{t}$

$\Rightarrow <v> = \frac{π R}{τ}$

$\Rightarrow <v> = \frac{π 160}{10}$ $= 16 π cm s^{- 1}$ .

$(b)$ Modulus of mean velocity vector is given by,

$|<\vec{v}>| = \frac{Modulus of displacement}{time duration}$

$\Rightarrow |<\vec{v}>| = \frac{|∆ \vec{r}|}{t}$

$\Rightarrow |<\vec{v}>| = \frac{2 R}{τ} . . . . . (1)$

$\Rightarrow |<\vec{v}>| = \frac{2 \times 160}{10}$ $= 32 cm s^{- 1}$ .

$(c)$ Let the point move from point $i$ to $f$ along the half-circle (as shown in the figure) and $v_{0}$ and $v$ be the speed at the points $i$ and $f$ respectively.
The tangential acceleration is given by,
$\frac{d v}{d t} = w_{t}$ (here, $w_{t}$ is the tangential acceleration)
As $w_{t}$ is constant, according to the problem,
So, $<v> = \frac{v_{0} + v}{2} . . . . . (2)$

So, using $(1)$ and $(2)$ ,
$\frac{v_{0} + v}{2} = \frac{π R}{τ}$

$\Rightarrow v_{0} + v = \frac{2 π R}{τ} . . . . . (3)$
Now the modulus of the mean vector of the total acceleration is,
$|< \vec{w} >| = \frac{∣ Δ \vec{v}]}{Δ t}$

$\Rightarrow |< \vec{w} >| = \frac{|\vec{v} - {\vec{v}}_{0}|}{τ}$

$\Rightarrow |< \vec{w} >| = \frac{v_{0} + v}{τ} . . . . . . (4)$ (As the direction of $\vec{v}$ and ${\vec{v}}_{0}$ are opposite.)
Using $(3)$ in $(4)$ , we get,
$|< \vec{w} >| = \frac{2 π R}{τ^{2}}$

Putting the values of $R$ and $τ$ , we get,

$|< \vec{w} >| = \frac{2 π \times 160}{10^{2}} = 3.2 π cm s^{- 2}$ .

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