10. V A red dot is painted on the tire of bicycle wheel of radius R. The wheel i
ID: 1660929 • Letter: 1
Question
10. V A red dot is painted on the tire of bicycle wheel of radius R. The wheel is rotating with a constant angular velocity w as shown below. For the questions below use the coordinate system given in the figure. fr Fig lc.pdif a) (5 pts) Write down expressions for the time dependence of the x and y coordinates of the dot for the case where the axle remains fixed on the y-axis. Use these to calculate the components of the velocity and acceleration of the dot. What is the magnitude of the acceleration? Is is what you would expect? Explain b) (5 pts) Now consider the case where the wheel is rolling down the negative x-axis without slipping. Assume the same value of w and repeat the calculations you made in part a). Does the magnitude of the dot's acceleration change? Is there any angular position(s) where the velocity of the dot is zero?Explanation / Answer
let initial position coordinates are (0,0)
acceleration due to gravity=-g
initial velocity in vector notation are (vx0,vy0)
at time t_0, the velocity is reversed.
let at time t_0, velocity in vector notation are (vx,vy)
at time t_0, horizontal velocity will be equal to initial horizontal velocity as horizontal acceleration=0
==>vx=vx0
at time t_0, vy=initial vertical velocty+acceleration*time
==>vy=vy0-g*t_0
now as velocity is reversed, new velocity=(-vx,-vy)=(-vx0,-vy0+g*t_0)
x component of position at time t_0=vx*t_0=vx0*t_0
y component of position at time t_0=vy0*t_0-0.5*g*t_0^2
t seconds after t_0, position x component =x’=x component of position at t_0+horizontal velocity*time
=vx0*t_0-vx0*t=vx0*(t_0-t)
which is equivalent of horizontal position at t_0-t.
vertical position t seconds after t_0=vertical position at t_0 + vertical velocity at t_0*time+0.5*acceleration*time^2
=vy0*t_0-0.5*g*t_0^2-(vy0-g*t_0)*t-0.5*g*t^2
=vy0*t_0-0.5*g*t_0^2-vy0*t+g*t_0*t-0.5*g*t^2
=vy0*(t_0-t)-0.5*g*(t_0-t)^2
which is equivalent of position at time t_0-t .
hence it is proved that the projectile will trace its trajectory and return to original position.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.