Physics 2210 Homework 18 Spring 2015

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Physics 2210 Homework 18 Spring 2015 Charles Jui April 12, 2015 IE Sphere Incline Wording A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with

1 Physics 2210 Homework 18 Spring 2015 Charles Jui April 12, 2015 IE Sphere Incline Wording A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle θ = 30.The sphere has mass M = 8 kg and radius R = 0.19 m. The coefficient of static friction between the sphere and the plane is µ = What is the magnitude of the frictional force on the sphere? See Figure 1. Figure 1: Sphere Incline Solution II.Newton law gives F net = M a 1

2 The net force is composed of the three forces F net = N + F g + f Since the sphere is moving along the incline, only its parallel component will be non-zero, i.e. F net = Mg sin θ f On the other hand, for rotation we have Taking the center as origin, we get 1 No slipping condition requires Combining the above results 2 τ net = I α τ net = Rf a = Rα Mg sin θ f M = F net M = R τnet I = R 2 f (2/5)MR 2 Solving for f f = 2 Mg sin θ 7 f = N Gymnast Wording A gymnast with mass m 1 = 41 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m 2 = 108 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam. See Figure 2. Figure 2: Gymnast 1 1 Only the frictional force gives non-zero contribution. 2 Moment of inertia for solid sphere is I = 2 5 MR2. 2

3 Figure 3: Gymnast 2 1. What is the force the left support exerts on the beam? 2. What is the force the right support exerts on the beam? 3. How much extra mass could the gymnast hold before the beam begins to tip? 4. Now the gymnast (not holding any additional mass) walks directly above the right support. What is the force the left support exerts on the beam? See Figure What is the force the right support exerts on the beam? 6. At what location does the gymnast need to stand to maximize the force on the right support? Solution Forces supports push up are F L and F R. Since the system is at rest, we got Since it s not rotating, it s true that 3 0 = F net = F L + F R m 1 g m 2 g 0 = τ net = L 6 F R L 6 F L + L 2 m 1g 1) Solving for the left force F L = 1 2 (4m 1 + m 2 )g F L = N 2) Solving for the right force F R = 1 2 (m 2 2m 1 )g F R = N 3 Taking the center of the beam as the origin. 3

4 3) The beam starts to tip when F R = 0, thus 0 = 1 2 (m 2 2m 1)g m 1 = m 2 /2 And so m 1 = m 1 m 1 = m 2 2 m 1 m 1 = 13 kg 4) Again we have 0 = F net = F R + F L m 1 g m 2 g Site of the force has changed, thus 4 0 = τ net = L 6 F R L 6 F L L 6 m 1g Thus the new left force F L = 1 2 (2m 1 + m 2 )g F L = N 5) Solving for the new right force F R = 1 2 m 2g F R = N 6) Hanging Beam Wording At the right edge of the beam A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of m b = 6.1 kg and the sign has a mass of m s = 16.8 kg. The length of the beam is L = 2.44 m. The sign is attached at the very end of the beam, but the horizontal wire holding up the beam is attached 2/3 of the way to the end of the beam. The angle the wire makes with the beam is θ = See Figure 4. 4 Origin at the beam s center. 4

Figure 4: Hanging Beam 1. What is the tension in the wire? 2. What is the net force the hinge exerts on the beam? 3. The maximum tension the wire can have without breaking is T = 977 N.

5 Figure 4: Hanging Beam 1. What is the tension in the wire? 2. What is the net force the hinge exerts on the beam? 3. The maximum tension the wire can have without breaking is T = 977 N. What is the maximum mass sign that can be hung from the beam? 4. What else could be done in order to be able to hold a heavier sign? Solution Since the beam is not moving, we have In coordinates 5 0 = F net = F b + T + F s + F h 0 = τ net = τ b + τ t + τ s + τ h 0 = T F hx (1) 0 = F hy m s g m b g (2) 0 = L 2 m bg cos θ 2L 3 T sin θ + Lm sg cos θ (3) 1) Solving for tension(from the Eq. (3)) T = 3(m b + 2m s )g 4 tan θ T = N 5 Taking the position of the hinge as the origin. 5

6 2) Components of the hinge force are F hx = T F hy = (m s + m b )g And thus its magnitude F h = (F hx ) 2 + (F hy ) 2 = T 2 + (m s + m b ) 2 g 2 or 9(mb + 2m s ) F h = g 2 16 tan 2 + (m s + m b ) θ 2 F h = N 3) Turning around the formula for the tension m s,max = 2T max tan θ 3g m s,max = kg m b 2 4) while still keeping it horizontal, attach the wire to the end of the beam keeping the wire attached at the same location on the beam, make the wire perpendicular to the beam attach the sign on the beam closer to the wall Meterstick Wording A meterstick (L = 1 m) has a mass of m = kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark. See Figure 5. Figure 5: Meterstick 1 1. What is the tension in the left string? 6

7 Figure 6: Meterstick 2 2. Now the right string is cut! What is the magnitude of the initial angular acceleration of the meterst about its pivot point? (You may assume the rod pivots about the left string, and the string remains vertical) 3. What is the tension in the left string right after the right string is cut? 4. After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. What is the angular speed when the meterstick is vertical? See Figure What is the magnitude of the acceleration of the center of mass of the meterstick when it is vertical? 6. What is the tension in the string when the meterstick is vertical? 7. Where is the angular acceleration of the meterstick a maximum? Solution 1) Since the stick is stable so 6 0 = F net = T L mg + T R 0 = τ net = (L/4)T R (L/4)T L Solving for the tension in the left string T L = 1 2 mg T L = N 2) The torque about the pivot is τ = lf sin φ = (L/4)mg 6 Origin is in the middle of the stick. 7

8 The second Newton law for rotation τ = Iα Moment of inertia of the rod about this pivot can be obtained by parallel axis theorem I = I 0 + md 2 = 1 12 ml2 + m(l/4) 2 = 7 12 ml2 So solving for the angular acceleration α = τ I = 12g 7L α = rad/s 2 3) At this moment, the acceleration of the center of mass is a = (L/4)α. So the second Newton law gives mg T L = F net = ma = m(l/4)α Solving for the tension T L = 4 7 mg T L = N 4) Using conservation of energy Solving for the angular speed Iω2 = mg L 4 mgl ω = = 2I 6g 7L ω = rad/s 5) Acceleration of the center of mass is completely centripetal 8 and hence a c = ω 2 l = 6g 7L L 4 a c = 3 14 g a c = m/s 2 7 Using the moment of inertia about the pivot - see above. 8 The radius of rotation being l = L/4. 8

6) Once again, II.Newton law tells us T mg = F net = ma c So the tension is and finally T = m(g + a c ) = m(g + (3/14)g) T = 17 14 mg T = 2.

9 6) Once again, II.Newton law tells us T mg = F net = ma c So the tension is and finally T = m(g + a c ) = m(g + (3/14)g) T = mg T = N 7) IE Sign Wording Right after the string is cut and the meterstick is still horizontal A sign has a mass of 1050 kg, a height h = 1 m, and a width W = 4 m. It is held by a light rod of length 5 m that is perpendicular to a rough wall. A guy wire at 23 to the horizontal holds the sign to the wall. Note that the distance from the left edge of the sign to the wall is 1 m. See Figure 7. Suppose we rely upon friction between the wall and the rod to hold up the sign (there is no hinge attaching the rod to the wall). What is the smallest value of the coefficient of friction µ such that the sign will remain in place? Figure 7: IE Sign 9

Solution The rod is not moving, so 0 = F net 0 = τ net The sign is exerting two equal 9 forces of (M/2)g at D and D W, where D is the length of the rod, while the wire pulls with tension T.

10 Solution The rod is not moving, so 0 = F net 0 = τ net The sign is exerting two equal 9 forces of (M/2)g at D and D W, where D is the length of the rod, while the wire pulls with tension T. Taking origin at the point of contact, our requirements translate into 10 0 = T sin θ (M/2)g (M/2)g + f 0 = N T cos θ 0 = D(M/2)g (D W )(M/2)g + DT sin θ And, of course, f = µn, where f is the force of friction and N is the normal force. Solving for the coefficient of static friction Ladder Wording µ = ( W 2D W µ = ) tan θ A ladder of length L = 2.6 m and mass m = 15 kg rests on a floor with coefficient of static friction µ s = Assume the wall is frictionless. See Figure 8. Figure 8: Ladder 1 1. What is the normal force the floor exerts on the ladder? 9 This can be explicitly shown, but but shoud be intuitively clear. 10 The x-axis points to the right, while the y-axis points upwards. That also makes the z-axis 10

11 Figure 9: Ladder 2 2. What is the minimum angle the ladder must make with the floor to not slip? 3. A person with mass M = 65 kg now stands at the very top of the ladder. What is the normal force the floor exerts on the ladder? See Figuer What is the minimum angle to keep the ladder from sliding? Solution There are four forces at play: Normal force from the floor : N 1 - pushing upwards Frictional force: f - pushing to the right Gravitational force (effectively) acting at the center of the ladder: mg - pointing down Normal force from the wall: N 2 - pushing to the left 1) For the ladder not to move, the net force needs to be zero, thus, in components 11 0 = f N 2 (4) 0 = N 1 mg (5) Thus we have N 1 = mg N 1 = N 11 The x-axis to the right, the y-axis upwards. 11

12 2) For the ladder not to rotate, the net torque needs to be zero 12, i.e. and so 0 = L 2 mg cos θ + LN 2 sin θ (6) tan θ = mg 2N 2 Using Eq. (4) with frictional force formula, we get N 2 = f = µn 1 = µmg so now ( ) 1 θ min = arctan 2µ θ min = ) At this, we introduced an additional force - weight of the person, which acts at the top end of the ladder. Thus, we need to modify the Eq. (5) to 0 = N 1 mg Mg which leads to N 1 = (m + M)g N 1 = N 4) Likewise, we need to modify Eq (6) to which yields while Eq. (4) is still applicable 0 = L 2 mg cos θ + LN 2 sin θ LMg cos θ tan θ = (m + 2M)g 2N 2 N 2 = f = µn 1 and thus θ min = arctan [( ) ] m + 2M 1 m + M 2µ θ min = With the origin at the bottom end of the ladder. 12

Physics 201. Professor P. Q. Hung. 311B, Physics Building. Physics 201 p. 1/1

Physics 201. Professor P. Q. Hung. 311B, Physics Building. Physics 201 p. 1/1 Physics 201 p. 1/1 Physics 201 Professor P. Q. Hung 311B, Physics Building Physics 201 p. 2/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia All elements inside the rigid