SOLUTION di x = y2 dm. rdv. m = a 2 bdx. = 2 3 rpab2. I x = 1 2 rp L0. b 4 a1 - x2 a 2 b. = 4 15 rpab4. Thus, I x = 2 5 mb2. Ans.

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Bennett Morton
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1 17 4. Determine the moment of inertia of the semiellipsoid with respect to the x axis and express the result in terms of the mass m of the semiellipsoid. The material has a constant density r. y x y a b 1 b x di x = y dm a m = Lv rdv a = rpb a1 - x L 0 a bdx = 3 rpab I x = 1 rp L0 = 4 15 rpab4 Thus, a b 4 a1 - x a b dx I x = 5 mb

2 Determine the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through point O. The slender rod has a mass of 10 kg and the sphere has a mass of 15 kg. O 450 mm Composite Parts: The pendulum can be subdivided into two segments as shown in Fig. a. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated. Moment of Inertia: The moment of inertia of the slender rod segment (1) and the sphere segment () about the axis passing through their center of mass can be computed from (I and (I G ) = G ) 1 = 1. The mass moment of inertia of 1 ml 5 mr each segment about an axis passing through point O can be determined using the parallel-axis theorem. 100 mm I O = I G + md = c 1 1 (10)(0.45 ) + 10(0.5 ) d + c 5 (15)(0.1 ) + 15(0.55 ) d = 5.7 kg # m

3 Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 00 mm O 0 kg>m. Composite Parts:The plate can be subdivided into two segments as shown in Fig. a. Since segment () is a hole, it should be considered as a negative part. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated. 00 mm Mass Moment of Inertia: The moment of inertia of segments (1) and () are computed as m 1 = p(0. )(0) = 0.8p kg and m = (0.)(0.)(0) = 0.8 kg. The moment of inertia of the plate about an axis perpendicular to the page and passing through point O for each segment can be determined using the parallel-axis theorem. I O = I G + md = c 1 (0.8p)(0. ) + 0.8p(0. ) d - c 1 1 (0.8)( ) + 0.8(0. ) d = kg # m

4 17 3. The mountain bike has a mass of 40 kg with center of mass at point G 1, while the rider has a mass of 60 kg with center of mass at point G. Determine the maximum deceleration when the brake is applied to the front wheel, without causing the rear wheel to leave the road. ssume that the front wheel does not slip. Neglect the mass of all the wheels. G 1 G 1.5 m 0.4 m Equations of Motion: Since the rear wheel is required to just leave the road, N = 0. Thus, the acceleration a of the bike can be obtained directly by writing the moment equation of motion about point. 0.4 m 0.4 m 0. m + M = (M k ) ; - 40(9.81)(0.4) - 60(9.81)(0.6) = -40a(0.4) - a = m>s = 5.61 m>s 60a(1.5)

5 Determine the shortest time possible for the rear-wheel drive, -Mg truck to achieve a speed of 16 m>s with a constant acceleration starting from rest. The coefficient of static friction between the wheels and the road surface is m s = 0.8. The front wheels are free to roll. Neglect the mass of the wheels. G 0.75 m m 1.5 m Equations of Motion: The maximum acceleration of the truck occurs when its rear wheels are on the verge of slipping. Thus, F = m s N = 0.8N. Referring to the free-body diagram of the truck shown in Fig. a, we can write : + F x = m(a G ) x ; 0.8N = 000a (1) + c F y = m(a G ) y ; N + N - 000(9.81) = 0 () + M G = 0; N (1.5) + 0.8N (0.75) - N () = 0 (3) Solving Eqs. (1), (), and (3) yields N = N N = N a = m>s Kinematics: Since the acceleration of the truck is constant, we can apply : + v = v 0 + at 16 = t t = 3.94 s

6 Determine the force developed in the links and the acceleration of the bar s mass center immediately after the cord fails. Neglect the mass of links and CD. The uniform bar has a mass of 0 kg m C D P 50 N 0.6 m Equations of Motion: Since the bar is still at rest at the instant the cord fails, v G = 0. Thus, (a G ) n = v G. Referring to the free-body diagram of the bar, Fig. a, r = 0 F n = m(a G ) n ; F t = m(a G ) t ; + M G = 0; T + T CD - 0(9.81) cos cos 45 = 0 0(9.81) sin sin 45 = 0(a G ) t T CD cos 45 (0.3) - T cos 45 (0.3) = 0 Solving, T = T CD = N = 51.7 N (a G ) t = m>s Since (a G ) n = 0, then a G = (a G ) t = 8.70 m>s R

7 The pendulum consists of a 10-kg uniform slender rod and a 15-kg sphere. If the pendulum is subjected to a torque of M = 50 N # m, and has an angular velocity of 3 rad>s when u = 45, determine the magnitude of the reactive force pin O exerts on the pendulum at this instant. O u M 50 N m 600 mm 100 mm Equations of Motion: Since the pendulum rotates about a fixed axis passing through point O, [(a G ) O ] t = a(r G ) O = a(0.3), [(a G ) ] t = a(r G ) = a(0.7), [(a, and [(a G ) ] n = v (r G ) = (3 G ) O ] n = v (r G ) O = (3 )(0.3) =.7 m>s )(0.7) = 6.3 m>s. The mass moment of inertia of the rod and sphere about their respective mass centers are (I G ) O = 1 and (I G ) =. Writing the moment equation of 5 mr = 1 ml = 1 1 (10)(0.6 ) = 0.3 kg # m 5 (15)(0.1 ) = 0.06 kg # m motion about point O, we have + M O = (M k ) O ; -10(9.81) cos 45 (0.3) - 15(9.81) cos 45 (0.7) - 50 = -10[a(0.3)](0.3) - 0.3a - 15[a(0.7)](0.7) a a = rad>s This result can also be obtained by applying M, where I O = I G + md O = I O a = 1. Thus, 1 (10)(0.6 ) + 10(0.3 ) + 5 (15)(0.1 ) + 15(0.7 ) = 8.61 kg # m + M O = I O a; -10(9.81) cos 45 (0.3) - 15(9.81) cos 45 (0.7) - 50 = -8.61a a = rad>s Using this result to write the force equations of motion along the n and t axes, F t = m(a G ) t ; 10(9.81) cos (9.81) cos 45 + O t = 10[16.68(0.3)] + 15[16.68(0.7)] O t = N F n = m(a G ) n ; O n - 10(9.81) sin 45-15(9.81) sin 45 = 10(.7) + 15(6.3) O n = 94.9 N Thus, F O = 4O t + O n = = N = 99 N

8 The two blocks and have a mass m and m, respectively, where m 7 m. If the pulley can be treated as a disk of mass M, determine the acceleration of block. Neglect the mass of the cord and any slipping on the pulley. r O a = ar c + M C = (M k ) C ; m g(r) - m g(r) = a 1 Mr ba + m r a + m r a a = a = g(m - m ) ra 1 M + m + m b g(m - m ) a 1 M + m + m b

The University of Melbourne Engineering Mechanics

The University of Melbourne Engineering Mechanics The University of Melbourne 436-291 Engineering Mechanics Tutorial Eleven Instantaneous Centre and General Motion Part A (Introductory) 1. (Problem 5/93 from Meriam and Kraige - Dynamics) For the instant