PHY : General Phyic CH 0 Workheet: Rotation Rotational Variable ) Write out the expreion for the average angular (ω avg ), in ter of the angular diplaceent (θ) and elaped tie ( t). ) Write out the expreion for the average angular acceleration (α avg ), in ter of the angular velocity (ω) and elaped tie ( t). 3) Write an expreion for the average angular velocity (ω avg ) in ter of initial and final angular velocitie (ω o and ω, repectively), where the angular acceleration i contant. 4) Cobine your anwer to the above exercie and derive the rotational kineatic equation in ter of the angular variable (θ,ω,α). ue the angular acceleration i contant. Rotational Motion 5) n ant i tanding on a oving CD, c fro the center. The CD i oving at 80 rp. a. When the CD ake a 45.0 o revolution, what i the length of the ant path? π θ=45 = 0.785 rad = θ r = 0.785 rad c =.57 c o 360 o n. b. What i the ant actual linear diplaceent ( r )? r o = c i + 0 c j o o r = r co 45 i + ( r in 45 ) j = (.4 c ) i + (.4 c ) j n. r = (.4 c - c ) i + (.4 c - 0 c ) j = (-0.59 c ) i + (.4 c ) j r = -0.59 c +.4 c =.53 c c. What i the rotational peed of the CD expreed in radian per ec? n. 80 rev π rad in ω = = 8.38 in rev 60 d. The ant ove to a new poition 4.0 c fro the center. i. What i the angular velocity of the ant copared to it peed at r=.0 c? n. The angular velocity i the ae, ω = 80 rad/. ii. What i the tangential peed of the ant copared to it peed at r =.0 c? n. Since r i doubled, the tangential peed v will alo double v 4 c = v c e. The ant then ove to the center of the CD. What i the angular velocity and tangential peed of the ant a the CD rotate? n. The angular velocity i the ae, ω = 80 rad/, but v = 0 / (tangential peed) rad
PHY : General Phyic CH 0 Workheet: Rotation Moent of nertia 6) Calculate the oent of inertia (in S unit) for the following object: a. kg a attached to a thin 0.5 long tring (a iple pendulu). n. = r = ( kg)(0.5 ) = 0.045 kg pendulu b. 0.5 kg hollow phere with a 0.0 diaeter. n. hollow phere = r = (0.5 kg)(0.05 ) = 0.005 kg 3 3 c. Two 0 kg olid phere (r=0.05 ) attached by a a-le 0.5 bar (iple dubbell), rotated about the iddle of the bar. n. 4 dubbell= r + L 5 4 dubbell= (0 kg)(0.05 ) + (0 kg)(0.30 ) 5 =.8 kg dubbell d. Two 0 kg olid phere (r=0.05 ) attached by a a-le 0.5 bar (iple dubbell), rotated about the center of one of the phere. n. dubbell= r + r + L 5 5 4 dubbell= (0 kg)(0.05 ) + (0 kg)(0.60 ) 5 = 3.6 kg dubbell e. eter tick (0.5 kg) rotated about it center of a. n. eter tick= L eter tick= (0.5 kg)(.0 ) = 0.05 kg eter tick Siple Pendulu Hollow Sphere Dubbell Dubbell Meter Stick Meter Stick f. eter tick (0.5 kg) rotated about one end. n. eter tick L eter tick= L + = L 3 eter tick= (0.5 kg)(.0 ) 3 = 0.050 kg
PHY : General Phyic 3 CH 0 Workheet: Rotation Torque: 7) Two people are itting on a 0 kg teeter-totter (L = 7.0 ), Peron (60 kg) and Peron B (30 kg). a. Peron it.5 fro the fulcru. What i the torque exerted by Peron on the teeter-totter (aue the a of the teeter-totter itelf can be neglected)? τ = r g = r g inθ k = n. τ = (-.5 )(60 kg)(-9.8 ) k = τ = 88 N k b. For the teeter-totter to be balanced, and in echanical equilibriu, with Peron B itting on the oppoing end, what torque ut Peron B apply to the teeter-totter? Draw a free body diagra of the -peron teetertotter yte. g B g τnet = τ + τb = 0 n. τ = - τ = - 88 N k B c. How far fro the fulcru ut Peron B it o that the teetertotter balance? τ = - g r k = 30 kg -9.8 r k = -88 N k n. ( ) B B B B r = B (-88 N ) ( 30 kg)( -9.8 ) i = 3 i d. Re-calculate all of the torque uing a fulcru located at one end of the balanced teetertotter.
PHY : General Phyic 4 CH 0 Workheet: Rotation inetic Energy & Rotation: 8) Deterine the rotational kinetic energy for the following object, rotated about the axi decribed above. a. a attached to a tring and rotating at 5 rad/ec. = ω = 0.045 kg 5 = 4.06 J rad n. b. hollow phere rotating about it center at 5 rad/ec. = ω = 0.005 kg 5 = 0.8 J rad n. c. Two 0 kg olid phere (r=0.05 ) attached by a ale 0.5 bar (iple dubbell), rotated about the iddle of the bar at 5 rad/ec. = ω =.8 kg 5 = 04.8 J rad n. d. Two 0 kg olid phere (r=0.05 ) attached by a ale 0.5 bar (iple dubbell), rotated about the center of one of the phere at 5 rad/ec. = ω = 3.6 kg 5 = 407.3 J rad n. e. eter tick (0.5 kg) rotated about it center of a, at 5 rp. = ω = 0.05 kg 0.54 = 0.007 J rad n. f. eter tick (0.5 kg) rotated about one end, at 5 rp. = ω = 0.0375 kg 0.54 = 0.0054 J rad n. 9) Deterine the rotational kinetic energy of a 0.5 kg golf ball, radiu of.0 c. a. Treat the golf ball a a olid phere that i rotating at 000 rp and iultaneouly traveling at 0 /, what i the rotational kinetic energy of the ball? n. -5 = r = (0.5 kg)(.00 ) =.4x0 kg 5 5 00 rev π rad in rad ω = =.05 in rev 60-5 rad -5 rot = ω = (.4x0 kg )(.05 ) =.3x0 J b. What i the total kinetic energy of the ball? -5 tot = tran + rot = v + ω = 0.5 kg 40 +.3x0 J 0 J n.
PHY : General Phyic 5 CH 0 Workheet: Rotation 0) Conider attached ae (Bug and E.) upended by a iple (olid cylindrical) pulley ( kg). Bug ha a a of 4 kg and E. ha a a of 8 kg. The radiu of the pulley i 0.. Starting fro ret, E. drop and Bug acend the ae aount. a. Draw a force vector diagra for each object (including the pulley). Note: the tenion force on each end of the rope are not equal. kg T T F N E Bug Pulley W Bug T T 8 kg W E W pulley b. pply Newton nd Law to E. and Bug. Ue it to olve for the tenion force, T and T. in ter of g, and Bug. n. F = g - T j = a ( E): Net F = T - g j = a (Bug): Net Bug Bug Bug (Pulley): F p= F N - T - T - pg j = 0 c. pply Newton nd Law (torque) to the pulley. τ = r T + r T = r T - r T k = α n. Net p p p p pr p a α r T - T = = = a p p rp rp 4 kg d. Ue the expreion for the tenion force in (b) and the torque equation obtained in (c) to olve for the acceleration of the yte (a). n. ( - Bug ) ( + Bug + p ) a = g = 3.0 e. What i the agnitude of the tenion force at each end of the rope, T and T, repectively? n. T = ( g - a ) = 54.4 N and T = Bug g + a = 5.8 N
PHY : General Phyic 6 CH 0 Workheet: Rotation f. What i E for E. and Bug, repectively, at the end of the diplaceent? n. = v = a r = 4. J = Bugv = Buga r Bug =.0 J g. What i the rotational E of the pulley at the end of the diplaceent? n. v = = =.46 4.6 J 8 kg v rot = ω = pr p = pv = 3.0 J r p 4 h. Calculate the total E and echanical energy of the yte at the end of the diplaceent. Copare thi value with the total E for the ae yte where the pulley i a-le. n. yte = + Bug + p = 4. J +.0 J + 3.0 J = 39. J For the ae yte w/ a-le pulley: ( - Bug) a = g = 3.6 ( + Bug) yte = + Bug = W + W Bug = ( + Bug ). a. r = 39. J