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A motion that is a combination of rotational and translational motion, e.g. a wheel rolling down the road. (a) same, (b) less. https://www.youtube.com/watch?v=vXNwYu1n9iw. Summary.

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## (a) same, (b) less

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**A motion that is a combination of rotational and**translational motion, e.g. a wheel rolling down the road.**(a) same, (b) less**https://www.youtube.com/watch?v=vXNwYu1n9iw**Summary**Rolling without slipping is a combination of translation and rotation where the point of contact is instantaneously at rest. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed v(r) = vcenter of mass The object will also move in a straight line in the absence of a net external force.**When an object experiences pure rotational motion about its**center of mass, all of its points move at right angles to the radius in a plane perpendicular to the axis of rotation with a speed proportional to the distance from the axis of rotation … v(r) = rω Thus points on opposite sides of the axis move in opposite directions, points on the axis do not move at all since r = 0 there … vcenter of mass = 0 and points on the outer edge move at the maximum speed … vouter edge = Rω**When an object experiences rolling motion the point of the**object in contact with the surface is instantaneously at rest … vpoint of contact = 0 and is the instantaneous axis of rotation. Thus, the center of mass of the object moves with speed … vcenter of mass = Rω and the point farthest from the point of contact moves with twice that speed vopposite the point of contact = 2vcom = 2Rω**Rolling as pure rotation about an axis through P**IP is the rotational inertia of the wheel about the axis through P. Substituting Eq. 11-4 into Eq. 11-3,**Differentiating Eq. 11-2 with**respect to time (with R held constant). and**Angular form of Newton’s second law**about an axis through the body’s center of mass as; Rolling down a Ramp cont.… (acom,x in the negative direction of the x axis) Less (consider the transfer of energy from rotation KE to gravitational PE)**(a) 0 (b) –x and (c) -y**(a) ± k (b) -i=k x j and (c) -j=k x -i**y**x (a) 1 and 3 tie, (b) 2 and 4 tie and (c) 5=0**(a) All tie (same torque, same t and thus same ∆L), (b)**Sphere, disk and hoop (reverse order of I)**Q. Calculate the angular momentum of a phonograph record**(LP) rotating at 331/3 rev/min. An LP has a radius of 15 cm and a mass of 150 g. A typical phonograph can accelerate an LP from rest to its final speed in 0.35 s, what average torque would be exerted on the LP? The angular momentum of a rotating body is L = I⍵. An LP is a solid disk. Consulting a table of moments of inertia, we find I = ½MR2. The angular velocity must be converted to rad/s ⍵ = 100/3 rev/min × 2π rad / rev × 1 min / 60 s = 3.4907 rad/s . Thus we find the angular momentum of the LP to be L = I⍵ = ½MR2⍵ = ½(0.15 kg)(0.15 m)2(3.4907 rad/s) = 5.8905 × 10-3 kg-m2/s . Torque is equal to the change in angular momentum with time τ = ΔL / Δt = (Lf - Li) / Δt = (5.8905 × 10-3 kg-m2/s - 0) / 0.35 s = 1.68 × 10-2 N-m .**In absence of external force**http://www.youtube.com/watch?v=0k276y9kuQQ**Examples**• decrease, (b) same (no net torques, therefore L is conserved) and (c) increases**from the figure that (with these positive direction choices)**its components are h**(since the initial velocity is purely horizontal) the**projectile motion equations become**To find the center of mass speed v on the plateau, we use**the projectile motion equations of Chapter 4. Eq. 4-21 (squared, and using d for the horizontal displacement)**Q. In a nightmare you dream that you are a hamster running**in an exercise wheel. Typical hamsters are 300 g and can run at speeds of 3.2 m/s. A typical exercise wheel has a moment of inertia about its center of 0.250 kg-m2. How fast should the wheel have been rotating in your dream? The radius of the wheel is 12.0 cm. Treat the hamster as a point mass. Hint what was the angular momentum of the system before the hamster started running? Before the hamster starts running, the exercise wheel is not rotating. Considered as a system, angular momentum must be conserved. Lf = Li . Here Lwheel + Lhamster = 0 . The wheel is a rotating object so its angular momentum is given by Lwheel = -I⍵ , where the minus sign indicates that it is into the paper. For a point particle, the angular momentum is Lhamster = Rmv out of the paper. Thus we have -I⍵ + Rmv = 0 . So the angular velocity of the wheel is ⍵ = Rmv / I = (0.3 kg)(0.12 m)(3.2 m/s) / (0.25 kg-m2/s) = 0.461 rad/s .

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