H: Rolling Motion - Bicycle Wheel Bookmark this page You have a bicycle wheel of radius R in non-slip rolling on a flat

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: H Rolling Motion Bicycle Wheel Bookmark This Page You Have A Bicycle Wheel Of Radius R In Non Slip Rolling On A Flat 1 (59.94 KiB) Viewed 10 times

H: Rolling Motion - Bicycle Wheel Bookmark this page You have a bicycle wheel of radius R in non-slip rolling on a flat horizontal surface. The center of the wheel is moving to the + direction with velocity ₁. The point at the bottom of the wheel at t = 0 is located at the origin; we will follow this special point. We will refer to the special point we are following as 'the point. The +y direction is chosen upward. The motion of the wheel is a translation at velocity added to a clockwise rotation of an origin-centered wheel at angular velocity w, raised upward so the bottom of the wheel has y = 0. The diagram provides a conceptual overview of the decomposition of the rolling motion explained in the next five problems of the sequence. אן || (F) What does the fact vector addition is commutative allow us? 1x + 1x G The goal is to determine the position, velocity, and acceleration of the point as a function of time. Before beginning this sequence, let's review. The 'Bicycle Wheel' sequence below is about finding the motion of a point on a wheel, the point initially at the bottom of the wheel, as a function of time, as the wheel rolls clockwise to the right on a horizontal surface. The motion is represented as the sum of three vectors. The vectors are a uniform circular motion on a circle centered at the origin, a lift by the displacement of the radius, and a translation to the right along the horizontal surface at the speed needed to simulate non-slip rolling along the surface.
Rotating Wheel at Origin 0/10 points (graded) Start with the wheel centered at the origin and rotating clockwise with angular velocity of magnitude w > 0. At t=0 the point has position (0, -R). Using the coordinates already specified, calculate the position vector R = (R(t), yr (t)) of the point on the rotating wheel (for the 'R' subscript think 'point on edge of rotating wheel at origin') as a function of time. Use the standard trigonometry convention for the direction of positive angle. Enter your responses in terms of some or all of v_x for , t fort, R for R, and omega for w. *R(t) - R. cos(w.t) Yr(t) = -R*sin(omega*t) -R.sin(w.t) SUBMIT You have used 2 of 10 attempts * Incorrect (0/10 points) -R*cos(omega*t) Translation of Lift 10/10 points (graded) Now add a constant vector displacement lift L = (L, YL) to position of the special point on the rotating wheel so that at time t = 0 the position of the special point is the origin. Determine the values of the lift displacement components. XL (t) 0 Enter your responses in terms of some or all of v_x for v₂,t fort, R YL (t) R SUBMIT 0 X R You have used 1 of 10 attempts for R, and omega for w. Save Save Show Answer
Linear and Angular Velocity in Rolling 10/10 points (graded) In lecture we derived a relationship between the angular velocity of a rolling wheel and the translational linear velocity of the rolling wheel. Choose the correct statements. The angular velocity is proportional to the linear velocity. The angular velocity is inversely proportional to the linear velocity. The angular velocity is the linear velocity times the radius of the wheel. The linear velocity is the radius of the wheel times the angular velocity. The radius of the wheel is the angular velocity times the linear velocity. SUBMIT You have used 1 of 2 attempts Correct (10/10 points) Translation Of Rolling 10/10 points (graded) *r(t) Now add a vector displacement translation r(t) = (xr(t), yr (t)) to the motion of the rotating wheel so that the center of the wheel moves to the + direction with velocity . Enter the values of the translation displacement below. Enter your responses in terms of some or all of v_x for U, t fort, R for R, and omega for w. v_x*t V₂ t YT (t) = 0 0 Save SUBMIT You have used 1 of 10 attempts Show Answer Save Show
Point Position 0.0/10.0 points (graded) Finally, the position of the point on the rolling bicycle wheel in table-top motion is Calculate the components of the position of the point. Enter your responses in terms of some or all of v_x for ₂, t fort, R for R, and omega for w. x(t) = = y(t) SUBMIT You have used 0 of 10 attempts Point Velocity 0.0/10.0 points (graded) Calculate the velocity of the point on the bicycle wheel (t) = n(t)+L(t) + #Ţ(t) = (x(t), y(t)). vy(t): SUBMIT You have used 0 of 10 attempts v(t) = di (t) dt Enter your responses in terms of some or all of v_x for , t fort, R for R, and omega for w. v₂ (t) = = (v₂ (t), vy(t)) =
Velocity Magnitude 0.0/10.0 points (graded) Calculate the magnitude of the velocity of the point on the bicycle wheel v(t) = |v(t)| Enter your responses in terms of some or all of v_x for , t fort, R for R, and omega for w. v(t) = SUBMIT You have used 0 of 10 attempts Zeros of Velocity 0.0/10.0 points (graded) Examine the expression you found for the magnitude of the velocity. Zeros of the velocity occur at all times that are Even multiples of 2TT □ Odd multiples of 2 Integer multiples of Integer multiples of Integer multiples of 2 SUBMIT You have used 0 of 3 attempts
Point 0.0/10.0 points (graded) Calculate the acceleration of the point on the bicycle wheel Acceleration ay(t) = a = and calculate the magnitude of the acceleration a = | al. Compare the magnitude of the acceleration of the rolling wheel to that of a rotating- only wheel (having the same radius and angular velocity). Enter your responses in terms of some or all of v_x for , t fort, R for R, and omega for w. az (t): SUBMIT You have used 0 of 10 attempts Velocity Magnitude - Simple Expression 0.0/10.0 points (graded) a(t) = d² x (t) dt² = (az (t), a, (t)) SUBMIT You have used 0 of 10 attempts Express the magnitude of the velocity of the point v(t) = v(t) in the form below. Complete by filling in the blank with an expression having no trig functions other than sine. The trigonometry identity cos(2x) = 1-2 sin² (x) will prove useful. Enter your responses in terms of some or all of v_x for ₂, t fort, R for R, and omega for w. v(t) = 2Rw( )1/2 Save Save
Path Length - One Revolution 0.0/10.0 points (graded) The point of the expression above is to have a simple form that is easily integrated. The length of the path the point takes during one revolution of the wheel is found by integrating the magnitude of the velocity to find the distance, = [² v where T = is the period of rotation of the rolling wheel. You can check the desired integral is 8 8 = SUBMIT You have used 0 of 10 attempts tw R 8 = 4Rw Enter the path length as a multiple of R by completing the expression. Enter your response as a number. v(t)dt, where we have doubled the value for half of the period in order to have just a sine function (rather than its absolute value) where the sine function is positive. To automatically evaluate the integral enter the following code into Mathematica (you can also enter the code at wolframalpha.com). T/2 wt sin()dt, 4* R * omega * Integrate[ Sin[omega*t/2], {t,0,Pi/omega} ] 116