Question

A car travels at$\mathbf{80}\mathbf{km}\mathbf{/}\mathbf{h}\mathbf{}$on a level road in the positive direction of an xaxis. Each tire has a diameter of$66\text{\hspace{0.17em}cm.}$Relative to a woman riding in the car, and in unit-vector notation,whatis the velocity$\overrightarrow{v}$at the (a) center (b) top, and (c) bottom of the tire? What is the magnitude aof the acceleration at the (d) center (e)top, and (f)the bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what is the velocity$\overrightarrow{v}$at the (g) centre (h) top, and (i) bottom of the tire? And the magnitude aof the acceleration at the (j) centre (k) top(l) bottom of each tire?

Short answer

Relative to woman:

a) Velocity at the center of the tire$=0\text{\hspace{0.17em}m/s}$

b) Velocity at the top of the tire$=(22\text{\hspace{0.17em}m/s})\widehat{i}$

c) Velocity at the bottom of the tire$=(-22\text{\hspace{0.17em}m/s})\widehat{i}$

d) Acceleration at the center of the tire$=0\text{\hspace{0.17em}m/s}$

e) Acceleration at the top of the tire$=(-1.5\times {10}^3{\text{\hspace{0.17em}m/s}}^{\text{2}})\widehat{j}$

f) Acceleration at the bottom of the tire $=(1.5\times {10}^3{\text{\hspace{0.17em}m/s}}^{\text{2}})\widehat{j}$

Relative to Hitchhiker

g) Velocity at the center of the tire$=(22\text{\hspace{0.17em}m/s})\widehat{i}$

h) Velocity at the top of the tire$=(44\text{\hspace{0.17em}m/s})\widehat{i}$

i) Velocity at the bottom of the tire$=0\text{m/s}$

j) Acceleration at the center of the tire$=\text{0m/s}$

k) Acceleration at the top of the tire$=(-1.5\times {10}^3{\text{\hspace{0.17em}m/s}}^{\text{2}})\widehat{j}$

l) Acceleration at the bottom of the tire$=(1.5\times {10}^3{\text{m/s}}^{\text{2}})\widehat{j}$

Step-by-step solution

Given

Velocity of the car, $v=80\text{\hspace{0.17em}km}/\text{hr}$

Diameter of the tire, $r=66\text{\hspace{0.17em}cm}$

To understand the concept

The rate of change of displacement with respect to time gives the velocity and the rate of change of velocity with respect to time is called acceleration.

Convert v and r into SI units

Before we go for calculation part, we will convert the$v\text{\hspace{0.17em}and}r$ into SI units.

$\begin{array}{c}v=80\frac{\text{km}}{\text{hr}}\\ =80\frac{\text{km}}{\text{hr}}\times \frac{1000\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}km}}\times \frac{1\text{\hspace{0.17em}hr}}{3600\text{\hspace{0.17em}s}}\\ =22.2\text{\hspace{0.17em}m}/\text{s}\end{array}$

$\begin{array}{c}r=66\text{cm}\\ =66\text{cm}\times \frac{0.01\text{\hspace{0.17em}m}}{1\text{cm}}\\ =0.66\text{\hspace{0.17em}m}\end{array}$

(a) Calculate the velocity   v→ at the center of the tire relative to the woman

In the cars reference frame, tire is performing only rotational motion. So center of the tire is at rest relative to the woman. So,$v_{\text{center}}=0\text{m/s}$

(b) Calculate the velocity  v→ at the top of the tire relative to the woman

Relative to woman, there is no translational motion of the tire, the motion is purely rotational.So the velocity of the top part of the tire is equal to the velocity which with any point of the tire is moving but it would be in the forward direction. The velocity of the point on the tire is determined in the above steps.

${\overrightarrow{v}}_{top}=(22\text{\hspace{0.17em}m/s})\widehat{i}$

(c) Calculate the velocity v→  at the bottom of the tire relative to the woman

The bottom point of the tire will have the same velocity as the top of the tire, only in the opposite direction

${\overrightarrow{v}}_{\text{bottom}}=(-22\text{m/s})\widehat{i}$

(d) Calculate the magnitude a of the acceleration at the center of each tirerelative to the woman

As the frame of reference is not accelerating, so the fixed points will have zero acceleration

${\overrightarrow{a}}_{\text{center}}=0{\text{m/s}}^{\text{2}}$

(e) Calculate the magnitude a of the acceleration at the top each tirerelative to the woman

In this the$\omega$(omega) is constant, which means the acceleration for points will be radial (centripetal),acting vertically downward.

,${\overrightarrow{a}}_{top}=-\frac{v^2}{r}\widehat{j}$

$\begin{array}{c}{\overrightarrow{a}}_{top}=-\frac{{(22\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s})}^2}{0.33\text{\hspace{0.17em}m}}\widehat{j}\\ =(-1.5\times {10}^3{\text{m/s}}^{\text{2}})\widehat{j}\end{array}$

(f) Calculate the magnitude a of the acceleration at the bottom of each tirerelative to the woman

The acceleration will be same as that of at the top, acting vertically upward.

$\begin{array}{c}{\overrightarrow{a}}_{bottom}=\frac{v^2}{r}\widehat{j}\\ =\frac{{(22\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s})}^2}{0.33\text{\hspace{0.17em}m}}\widehat{j}\\ =(1.5\times {10}^3{\text{m/s}}^{\text{2}})\widehat{j}\end{array}$

(g) Calculate the velocity  v→ at the center of the tire relative to the hitchhiker

Now, the car is having both translational and rotational motion,

${\text{v}}_{\text{center}}=22\text{m/s}$

.${\overrightarrow{v}}_{\text{center}}=(22\text{m/s})\widehat{i}$

(h) Calculate the velocity  v→ at the top of the tire relative to the hitchhiker

$\begin{array}{c}{v}_{top}=v_{rotational}+v_{translational}\\ =22\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s}+22\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s}\\ =+44\text{m/s}\\ {\overrightarrow{v}}_{\text{top}}=(+44\text{m/s})\widehat{i}\end{array}$

(i) Calculate the velocity v→ at the bottom of the tire relative to the hitchhiker

The bottom of the tire is in firm contact with the road at any instant, so we have

$\begin{array}{c}{v}_{bottom}=v_{rotational}+v_{translational}\\ =-22\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s}+22\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s}\\ =0\text{m/s}\end{array}$

(j) Calculate the magnitude a of the acceleration at the centre of each tire relative to the hitchhiker

As the frame of reference is not accelerating, so the fixed points will have zero acceleration

$a_{\text{center}}=0{\text{\hspace{0.17em}m/s}}^{\text{2}}$

(k) Calculate the magnitude a of the acceleration at the top each tire relative to the hitchhiker

As we are translating with the constant velocity, so translational acceleration will be zero, only rotational acceleration will be there,

$\begin{array}{c}{\overrightarrow{a}}_{top}=-\frac{v^2}{r}\widehat{j}\\ =-\frac{{(22\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s})}^2}{0.33\text{\hspace{0.17em}m}}\widehat{j}\\ =-(1.5\times {10}^3{\text{m/s}}^{\text{2}})\widehat{j}\end{array}$

(l) Calculate the magnitude a of the acceleration at the bottom of each tire relative to the hitchhiker

The acceleration will be same as that of at the top,

$\begin{array}{c}{\overrightarrow{a}}_{bottom}=\frac{v^2}{r}\widehat{j}\\ =\frac{{(22\text{\hspace{0.17em}\hspace{0.17em}m}/\text{s})}^2}{0.33\text{\hspace{0.17em}m}}\widehat{j}\\ =(1.5\times {10}^3{\text{m/s}}^{\text{2}})\widehat{j}\end{array}$

The acceleration at the bottom of the tire, relative to the hitchhiker is $(1.5\times {10}^3{\text{m/s}}^{\text{2}})\widehat{j}$.