Question

A car travels at

80 km h^{- 1}

on a level road in the positive direction of an x axis. Each tire has a diameter of

66 cm

. Relative to a woman riding in the car and in unit-vector notation, what are the velocity

\vec{v}

at the (a) center, (b) top, and (c) bottom of the tire and the magnitude

a

of the acceleration at the (d) center, (e) top, and (f) bottom of each tire? Relative to a hitchhiker sitting next to the road and in unit-vector notation, what are the velocity v at the (g) center,
(h) top, and (i) bottom of the tire and the magnitude

a

of the acceleration at the (j) center, (k) top, and (1) bottom of each tire?

Accepted Answer

The velocity of the car is a fixed quantity

$\vec{v} = (80 km h^{- 1}) (1000 m) (3600 s) \hat{i} = (+ 22 m s^{- 1}) \hat{i}$
and the radius of the wheel is $r = \frac{0.66}{2} = 0.33 m$

(a) In the car's point of frame (where the lady perceives herself to be at rest) the road is moving backward at ${\vec{v}}_{rood} = - v = - 22 m s^{- 1}$ , and the motion of the tire is purely rotational.

In this frame of reference, the center of the tire is "fixed" so $v_{center} = 0$ .

(b) Since the tire's motion is just rotational (not translational) in this frame, ${\vec{v}}_{top} = (+ 22 m s^{- 1}) \hat{i}$ .

(c) The bottom-most point of the tire is (for a instant) in firm contact with the road (not skidding) and has the same velocity as the road: ${\vec{v}}_{bottom} = (- 22 m s^{- 1}) \hat{i}$ .

(d) This point of reference is not accelerating, so "fixed" points within it have zero acceleration; thus, $a_{center} = 0$

(e) Not only is the motion is just rotational in this frame, but we also have $ω =$ $constant$ , which implies the only acceleration for points on the rim is radial (centripetal). Therefore, the magnitude of the acceleration is
$a_{top} = \frac{v^{2}}{r} = \frac{(22 m s^{- 1})^{2}}{0.33 m} = 1.5 \times 10^{3} m s^{- 2}$

(f) The magnitude of the acceleration is the same as in part (d): $a_{bottom} = 1.5 \times 10^{3} m s^{- 2}$ .

(g) Now, we test the situation in the road's frame of reference (where the road is "fixed" and it is the car that appears to be moving). The center of the tire goes through translational motion only, while points at the rim goes through a combination of translational and rotational motions. The velocity of the center of the tire is $\vec{v} = (+ 22 m s^{- 1}) \hat{i}$
(h) In part (b), we found ${\vec{v}}_{wop. car} = + v$
${\vec{v}}_{top. ground} = {\vec{v}}_{top, car} + {\vec{v}}_{car, ground} = v \hat{i} + v \hat{i} = 2 v \hat{i}$

which gives $2 v = + 44 m s^{- 1}$ .

(i) We can begin as in part (h) or simply remind that the bottom-most point is in firm contact with the (zero-velocity) road. Either way, the answer is zero.

(j) The translational motion of the center is fixed or ay constant; it does not accelerate.

(k) Since we are equating between constant-velocity frames of reference, the accelerations are not getting affected. The answer is as it was in part (e): $1.5 \times 10^{3} m s^{- 2}$ .

(1) As explained in part (k), $a = 1.5 \times 10^{3} m s^{- 2}$ .

Resnick & Halliday Solutions for Chapter: Rolling, Torque, and Angular Momentum, Exercise 1: Problems

Resnick & Halliday Physics Solutions for Exercise - Resnick & Halliday Solutions for Chapter: Rolling, Torque, and Angular Momentum, Exercise 1: Problems

Questions from Resnick & Halliday Solutions for Chapter: Rolling, Torque, and Angular Momentum, Exercise 1: Problems with Hints & Solutions

Learn and Prepare for any exam you want !