Phys exam 4 inc 10-25 angular momentum

Question 1. C Suppose unseat he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s (18.5 rads/s) what average torque was entered if this takes 12.0 seconds T =

-.769 N*m Solve by T (torque) = I (moment of inertia) * alpha (angular acceleration) w final = w initial + alpha * t alpha = (w final - w initial) / t w initial = 18.85 rads / a w final = 0 rad/s (comes to a stop) alpha = 0-18.85 / 12 = -1.57 rad/s^2 T = I * alpha T = .4900 kgm^2 (-1.57 rad/s^2) - .769 N*m

Question 1. A Calculate the angular momentum of an ice skater spinning at 6.00 rev/s (37.7 rad/s) given his moment of inertia is .4900kg m^2 L =

18.473 kgm^2/s Solve by L = I * w (.4900kgm^2 )(37.7 rad/s)

Question 1. B He reduces his rate of spin (his angular velocity by extending his arms and increasing his moment of inertia. Find the value of his new moment of inertia of his angular velocity drops to 1.00 rev / s I =

2.9400 kgm^2 Solve by L initial = L final I initial * w initial = I final * w final w final = 1.00 revs / s x 2pi / rev = .6283 rad / s L initial = 18.473 kgm^2/s L initial = L final L initial * w initial = I final * w final 18.473 = I final * 6.283 I final = 2.94 kgm^2

Question 5. B Which three of the following are physics principles that support your answer? Rotational inertia cannot be changed without changing the mass in the system. Angular momentum is conserved. Your rotational inertia increases as you crawl outward. Angular momentum is rotational inertia times rotational velocity. The rate of rotation is constant when there are no external forces.

Angular momentum is conserved. Your rotational inertia increases as you crawl outward. Angular momentum is rotational inertia times rotational velocity.

Question 5. A You sit at the middle of a large turntable at an amusement park as it is set spinning and then allowed to spin freely. When you crawl toward the edge of the turntable, does the rate of the rotation increase, decrease, or remain unchanged? The rate of rotation decreases. The rate of rotation remains unchanged. The rate of rotation increases.

The rate of rotation decreases.

question 6. C A child is swinging a 32 g yo-yo around in a horizontal circle at a frequency of 3 Hz. (c) What is the frequency of rotation? Hz

frequency and tangential speed v = 2 pi r f f' is the new freq f' = v' / 2 pi r ' f ' = (19.32 m/s) / 2 pi (.52 ) = 5.80 Hz

Question 6 A. A child is swinging a 32 g yo-yo around in a horizontal circle at a frequency of 3 Hz. (a) If the string is 73 cm long, what is the tangential speed of the yo-yo? m/s

mass of yo yo = .032 kg freq of rotation = 3 Hz radius of circle r = .73 m tangential speed v v = 2 pi r f v = 2 pi (.73m) (3Hz) v = 13.76 m/s

Question 6. B A child is swinging a 32 g yo-yo around in a horizontal circle at a frequency of 3 Hz. The girl suddenly pulls in the string so that the radius of the circle decreases to 52 cm. (b) What is the new tangential speed of the yo-yo?Hint: Think of conservation of angular momentum. m/sec(c) What is the frequency of rotation? Hz

r = .52 m anglular momentum is conserved L = Iw = mr^2w I = mr^2 is the moment of inertia of the yo yo w is the angular velocity mr^2w = mr'^2 w' r ' = new angular velocity w ' = new radius rv = r' v' v' =( r / r' ) v (.73 / .52 ) * 13.76 m/s v' = 19.32 m/s

Question 3 Suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. As you crawl toward the edge, the angular momentum of you and the turntable increases. decreases. remains the same, but the RPMs decrease. none of these decreases in direct proportion to your decrease in RPMs.

remains the same, but the RPMs decrease.

Question 4 A 2.2 m radius playground merry-go-round has a rotating mass of 120 kg and is rotating with an angular velocity of 0.550 rev/s.What is its angular velocity after a 15.0 kg child gets onto it by grabbing its outer edge? The added child is initially at rest.Treat the merry-go-round as a solid disk (I = 12M R2) , and treat the child as a point mass (I = M R2)

step 1 : find moment of inertia of the merry go round step 2: find moment of inertia of the child step 3: final initial angular momentum step 4: find final angular momentum to solve to final angular velocity moment of inertia of the merry go round I mr = 1/2 M mr R^2 I mr = 1/2 (120kg) (2.2m)^2 I mr = 290.4 kh m^2 moment of inertia of the child I child = M child R^2 I child = (15 kg) (2.2m)^2 I child = 72.6 kg m^2 initial angular momentum Li = I mr Wi Li = (290.4kgm^2) (.550 rev/s) Li = 159.72 kgm^2 /s final angular momentum (after the child grabs the merry go round) Lf = (I mr + I child) wf Li = Lf 159.72 khm^2 /s = (290.4 + 72.6) wf wf = .440 rev/s final angular velocity .44 rev/s

Question 2 We studied three conserved quantities, energy, momentum and angular momentumFor energy we have the high level 'Energy Balance' formula: ΔEtotal = FΔx cos(𝜃) Where F is the magnitude of the constant force, and Δx is the magnitude of the displacement. For momentum we have the high level 'Momentum Balance' formula: Δp total = F avgΔt Use the technique of translating to rotational variables to determine the high-level balance formula for angular momentum: ΔLtotal = 𝜃avgΔt ΔLtotal = tavgΔ𝜏 ΔLtotal = 𝛼avgΔ𝜏 ΔLtotal = 𝜏avgtavg ΔLtotal = 𝜏avgΔt ΔLtotal = 𝜔avgΔt

ΔLtotal = 𝜏avgΔt