The 2000 kg mass of a car includes fourtires, each of mass (including wheels) 47
ID: 1744263 • Letter: T
Question
The 2000 kg mass of a car includes fourtires, each of mass (including wheels) 47kg and diameter 0.90 m. Assume each tireand wheel combination acts as a solid cylinder. (a) Determine the total kinetic energy of thecar when traveling 90 km/h.1 J
(b) Determine the fraction of the kinetic energy in the tires andwheels.
2%
(c) If the car is initially at rest and is then pulled by a towtruck with a force of 2000 N, what is theacceleration of the car? Ignore frictional losses.
3 m/s2
(d) What percent error would you make in part (c) if you ignoredthe rotational inertia of the tires and wheels?
4% (a) Determine the total kinetic energy of thecar when traveling 90 km/h.
1 J
(b) Determine the fraction of the kinetic energy in the tires andwheels.
2%
(c) If the car is initially at rest and is then pulled by a towtruck with a force of 2000 N, what is theacceleration of the car? Ignore frictional losses.
3 m/s2
(d) What percent error would you make in part (c) if you ignoredthe rotational inertia of the tires and wheels?
4%
Explanation / Answer
The mass of the car including four tires is M = 2000kg The mass of each wheel is m = 47 kg The diameter of each wheel is d = 0.90 m (a)The total kinetic energy of the car is K = (1/2)Mv2 ------------------(1) Here,v = 90 km/h = 90 * (5/18) m/s = 25 m/s or K = (1/2) * 2000 * (25)2 or K = 625 * 103 J = 625 kJ (b)Each tire and wheel combination acts as a solidcylinder.The moment of inertia of the solid cylinder is I = (1/2)mr2 = (1/2)m * (d/2)2 = (1/8)m* d2 The kinetic energy in the tires and wheels is K1 = 4 * (1/2)Iw2 Here,v = r * w or w = (v/r) = (v/d/2) = (2v/d) or K1 = 4 * (1/2) * (1/8)m * d2 *(2v/d)2 or K1 = mv2---------------------(2) Dividing equation (2) by equation (1),we get (K1/K) = (mv2/(1/2)Mv2) or (K1/K) = (2m/M) = (2 * 47/2000) = 47 *10-3 = 0.047 or K1= 0.047 * K Therefore,the fraction of the total kinetic energy in thetires and wheels is 0.047 (c)The car is initially at rest,that is,u = 0 m/s The car is then pulled by a tow truck with a force of F = 2000N Let the distance travelled by the car when it is pulled beS,therefore,we get F * S = K or S = (K/F) or S = (625 * 103/2000) = 312.5 m Let the acceleration of the car during this time bea,therefore,we get v2 - u2= 2aS or a = (v2 - u2/2S) or a = ((25)2 - (0)2/2 * 312.5) = 1m/s2 (d)When the rotational inertia of the tires and wheels isignored,then the kinetic energy is (K - K1),therefore,weget F * S = (K - K1) or S = [(K - K1)/F] or S = [((1/2)Mv2 -(1/2)(4m)v2)/F] or S = [(1/2) * v2* (M - m)/F] or S = [(1/2) * (25)2 * (2000 - 4 * 47)/2000] or S = 283.1 m Let the acceleration of the car during this time bea1,therefore,we get v2 - u2= 2a1S or a1= (v2 - u2/2S) or a1= ((25)2 - (0)2/2 *283.1) or a1= 1.1 m/s2 Therefore,the percent error we would make in part (c) if weignored the rotational inertia of the tires and wheels (a/a1) * 100 = (1/1.1) * 100 = 90.9 % (d)When the rotational inertia of the tires and wheels isignored,then the kinetic energy is (K - K1),therefore,weget F * S = (K - K1) or S = [(K - K1)/F] or S = [((1/2)Mv2 -(1/2)(4m)v2)/F] or S = [(1/2) * v2* (M - m)/F] or S = [(1/2) * (25)2 * (2000 - 4 * 47)/2000] or S = 283.1 m Let the acceleration of the car during this time bea1,therefore,we get v2 - u2= 2a1S or a1= (v2 - u2/2S) or a1= ((25)2 - (0)2/2 *283.1) or a1= 1.1 m/s2 Therefore,the percent error we would make in part (c) if weignored the rotational inertia of the tires and wheels (a/a1) * 100 = (1/1.1) * 100 = 90.9 % Therefore,the percent error we would make in part (c) if weignored the rotational inertia of the tires and wheels (a/a1) * 100 = (1/1.1) * 100 = 90.9 %Related Questions
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