 # A bicycle with 0.80-m-diameter tires is coasting on a level road at 5.6 m/s. A s Cynthia Bell 2021-12-16 Answered

A bicycle with 0.80-m-diameter tires is coasting on a level road at 5.6 m/s. A small blue dot has been painted on the tread of the rear tire. a. What is the angular speed of the tires? b. What is the speed of the blue dot when it is 0.80 m above the road? c. What is the speed of the blue dot when it is 0.40 m above the road?

You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Melinda McCombs

Given values:
$2r=0.8m$
$r=0.4m$
${x}_{1}=0.8m$
${\upsilon }_{cm}=5.6\frac{m}{s}$
a) We will find the angular speed as:
$\omega =\frac{{\upsilon }_{cm}}{r}$
$\omega =\frac{5.6\frac{m}{s}}{0.40m}$
$\omega =14\frac{rad}{s}$
The angular velocity of a point at height ${x}_{1}$ will be the sum of the line velocity and the velocity of the center of mass of the wheel.
$\upsilon ={\upsilon }_{cm}+r\omega$
$\upsilon =5.6\frac{m}{s}+0.4m\cdot 14\frac{rad}{s}$
$\upsilon =11.2\frac{m}{s}$
Using the Pythagorean theorem, we have:
${\upsilon }^{2}={\left(r\omega \right)}^{2}+{\upsilon }_{c{m}^{2}}$
$\upsilon =\sqrt{{\left(r\omega \right)}^{2}+{\upsilon }_{cm}^{2}}$
$\upsilon =\sqrt{{\left(0.4m\cdot 14\frac{rad}{s}\right)}^{2}+{\left(5.6\frac{m}{s}\right)}^{2}}$
$\upsilon =7.9195\frac{m}{s}$
a) $\omega =14\frac{rad}{s}$
b) $\upsilon =11.2\frac{m}{s}$
c) $\upsilon =7.9195\frac{m}{s}$

We have step-by-step solutions for your answer! Jim Hunt

(1) Angular speed $=\frac{5.6}{0.4}=14\frac{radians}{\mathrm{sec}}$, where 0.4m is the radius of the wheel.
Since 2n radians is one revolution, $14\frac{radians}{s}$ is $\frac{14}{2}\pi =2.23rev\frac{s}{s}$ or 133.7 rpm.
(2) We have to add the tangential speed to the translational speed. The direction is important so we should consider velocity rather than the speed because the tangential velocity is changing but the translational velocity stays the same. At 0.80m the blue dot is at its highest point and the tangential velocity is in the direction of $motion=14×0.4=5.6\frac{m}{s}$. So the speed of the blue dot is $5.6+5.6=11.2\frac{m}{s}$.
(3) The blue dot is 0.4m above the road in two different places. The tangential velocity is vertically upwards or vertically downwards, for the trailing or leading edge of the wheel respectively. The resultant velocity is at an angle of 45° to the road and has a magnitude (speed) of $5.6\sqrt{2}=7.92\frac{m}{s}$. This magnitude applies equally to both endpoints of the horizontal diameter. The direction for each velocity is different.

We have step-by-step solutions for your answer! Don Sumner

a) d = diameter of tire = 0.80 m
r= radius of tire = (0.5) d = (0.5) (0.80) = 0.40 m
v = speed of bicycle = 5.6 m/s
w = angular speed of the tire
Speed of cycle is given as
$v=rw$
$5.6=\left(0.40\right)w$

b)
${V}^{\prime }=$ speed of blue dotask
Speed blue of dot is given as
${v}^{\prime }=v+rw$
${v}^{\prime }=5.6+\left(0.40\right)\left(14\right)$

C. here angle b/w both the vectors will be 90 degrees,
$R=\left[{v}^{2}+\left(rw{\right)}^{2}{\right]}^{0.5}$
$R=\left[2\ast {5.6}^{2}{\right]}^{0.5}$

We have step-by-step solutions for your answer!