Dynamics Solutions

Dynamics 101 – Conservation of Energy of a Particle As the climber ascends the rock wall he loops the safety rope through carabiners anchored to the wall to limit the distance that he will drop in the event of a fall. A 180-lb climber is a distance d above the last anchor point as shown below. The maximum force that the climber can sustain from his harness without serious injury is 800lb. Find the maximum safe distance d assuming that the rope behaves like a spring with a stiffness k=25lb/ft. Solution: If the climber falls, he will drop a total height h=2d+s, where s is the stretch of the rope, as shown below. Since the velocity of the climber is zero in both the start and end position, the potential energy is the same for both. The stretch associated with the maximum load of 800lb can be found from We can now calculate the distance d as Author: Kennedy, T. C. Subject: Dynamics Topic: Conservation of Energy of a Particle Problem Length: 10 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:  Dynamics 102 – Particle Kinetics with Polar Coordinates Does Spiderman violate the laws of dynamics? In one of his movies, he keeps pace with a speeding police car by swinging like a pendulum on the end of a line of spider silk of length L as shown below. Find the average horizontal velocity using this means of transport. Solution: We begin by drawing a free body diagram of Spiderman. We write Newton’s second law for the θ coordinate. We can rewrite this result as where Next, we use the chain rule of calculus to rewrite this again as This is now in a form where we can separate variables and integrate. Solving for ω and setting ω=dθ/dt gives Again, we separate variables and integrate over one 180° arc. Solving for the time to complete the arc gives. The horizontal distance traveled during this time is 2L. Therefore, the average horizontal velocity is given as From the movie, it appears that L~150ft. This corresponds to a velocity of 38.7ft/s=26.4mph. Is this sufficient to keep pace with a speeding police car? Author: Kennedy, T. C. Subject: Dynamics Topic: Particle Kinetics with Polar Coordinates Problem Length: 20 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:  Dynamics 103 – Particle Impulse and Momentum Does the superhero Hancock violate the laws of dynamics? In the movie, he brings a moving train to an abrupt stop using his bare hands. Find the force required to bring a 200ton locomotive traveling at 20mph to a complete stop in one second as shown below. Solution: We begin by writing the impulse-momentum relation for the locomotive subjected to a force in the x-direction. To make the integration simple, let’s assume that the force Fx is constant. This gives For this case we have m=12,422slugs, vx-start=-29.3ft/s, and t=1s. This gives Fx=364,000lb. Let’s now draw a free body diagram of Hancock, assuming he weighs 200lb. The horizontal force of 364,000lb must be transmitted into the ground to keep Hancock motionless as shown in the movie. If this were done by friction, it would require μN=364,000lb. This results in a coefficient of friction μ=1820. This is not possible. Perhaps he braces his feet against a railroad tie. Can a 6inX10inX72in wooden beam carry a 364,000lb force? Author: Kennedy, T. C. Subject: Dynamics Topic: Particle Impulse and Momentum Problem Length: 10 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other: Dynamics 104 – Rigid Body Kinetics without Rotation Why do the front brakes on a car wear out faster than the rear brakes? To get some insight into this, consider the following problem. The car shown below skids to a stop with the brakes locked. The coefficient of friction between the tires and the road is 0.9. Compare the stopping force on the front tires to those on the rear. Solution: We begin by drawing a free body diagram of the car with a coordinate system located at the bottom of the rear wheels. Applying force equilibrium gives Solving gives acx=-μg and Nr=mg-Nf. Now apply moment equilibrium with the recognition that α=0, aox=acx, and aoy=0. This gives and Now let’s form the ratio of the front friction force to the rear friction force. This means that stopping force from the front is almost seven times that of the rear for this case. Author: Kennedy, T. C. Subject: Dynamics Topic: Rigid Body Kinetics without Rotation Problem Length: 15 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:  Dynamics 105 – Rigid Body Kinetics with Rotation A billiard ball with radius r=2.25in is struck in such a way that it is propelled to the right with center velocity vco=10ft/s and backspin (counterclockwise rotation) of ωo=20radians/s. Find the distance that the ball travels before its rotation becomes clockwise if the coefficient of friction is μ=0.5. Solution: We begin by drawing a free body diagram of the ball. Force equilibrium gives From these we get and Integrating this with respect to time and using the fact that the initial velocity is vo gives Integrating this with respect to time and using the fact that the initial position is xc=0 gives Moment equilibrium gives From this we get Integrating this with respect to time and using the fact that the initial angular velocity is ωo gives The rotation direction changes from counterclockwise to clockwise when ω=0. Setting ω=0 and solving for t gives Substituting this value for time in to the expression for position xc gives Author: Kennedy, T. C. Subject: Dynamics Topic: Rigid Body Kinetics with Rotation Problem Length: 20 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:  Dynamics 106 – Particle Kinetics with Normal and Tangential Coordinates A 1500-lb race car travels around a curve with a radius ρ=400ft at constant speed. Wings on the car create a down force equal to cv2 where c=0.02slug/ft and v is the speed in ft/s. If the coefficient of friction between the tires and the road is μ=1.5, find the maximum speed that the car can have without slipping. Solution: We begin by drawing a free body diagram of the car assuming that it is on the verge of slipping. Since the car is traveling with constant speed (i.e., ), Newton’s second law gives This gives and Solving for v gives Author: Kennedy, T. C. Subject: Dynamics Topic: Particle Kinetics with Normal and Tangential Coordinates Problem Length: 10 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:  Dynamics 107 –Kinetics of a Particle A 0.5kg block starts from position 1 with velocity v1at the top of the disk and slides around the smooth disk with radius r=0.2m. Find the maximum allowable value for v1 such that the block does not lose contact with the disk when it passes position 2 at the bottom of the disk. The spring has a stiffness k=400Newton/m and an unstretched length of 0.1m. Solution: We will begin by analyzing the motion from point 1 to point 2 using conservation of energy since the normal force between the block and the disk does no work. We also observe that there is no change in potential energy in the spring. Now we draw a free body diagram of the block in position 2. We assume that the block is on the verge of losing contact with the disk so that N=0. Newton’s second law gives Substituting the expression for v2 gives Solving for v1 gives Author: Kennedy, T. C. Subject: Dynamics Topic: Kinetics of a Particle Problem Length: 15 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:  Dynamics 108 – Kinematics of Rigid Bodies Point A moves to the left at a speed of 4ft/s. Point B moves to the right at a speed of 2ft/s. Find the speed of point C at the instant shown. Let’s write the velocity of point C using point A as a reference. Let’s write the velocity of point C using point B as a reference. Equating the two expressions for gives From this we get and We can solve these to get Substituting this into the expression for gives and Author: Kennedy, T. C. Subject: Dynamics Topic: Kinematics of Rigid Bodies Problem Length: 15 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other: Dynamics 109 – Kinematics of Rigid Bodies A ring with outer radius r4=2.5in and inner radius r3=2in rolls clockwise without slipping at ωR=10radians/s. The center shaft with radius r1=1in rotates counterclockwise at ωS =5radians/s. Find the angular velocity of the bearings ωB with radius r2=0.5in. There is no slipping at the contact points. Solution: For a wheel that rolls without slipping, there is a simple relation between the velocity of the center of the wheel and its angular velocity; i.e., vc=-r4ωR. Using the unit vectors in the figure above, we can write The velocity of point b is The velocity of point a is The velocity of point d can be written as However, because point d does not slip, we can set From this we can solve for ωB as Author: Kennedy, T. C. Subject: Dynamics Topic: Kinematics of Rigid Bodies Problem Length: 20 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:  Dynamics 110 – Kinetics of Rigid Bodies with Rotation A bar with length L is released from the position shown with θ=30°. Find the acceleration of the contact point if the surface is frictionless. Solution: We begin by drawing a free body diagram of the bar. Considering force equilibrium in the x-direction gives Since ω=0 for the initial position, the accelerations of points c and o are related through the rigid-body kinematic relation Combining this with the previous equation, we conclude that Considering moment equilibrium about point o gives Combining this equation with the previous one gives Author: Kennedy, T. C. Subject: Dynamics Topic: Kinetics of Rigid Bodies with Rotation Problem Length: 15 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other:  Dynamics 111 – Kinetics of Particles with Polar Coordinates While Cleetus is fishing, a game warden comes along and gives Cleetus a citation because one of his fish is 2% below the minimum legal weight. Cleetus tells the game warden that the fish is legal because the game warden’s fish scale does not measure the true weight because it does not account for the acceleration that the fish experiences due to the rotation of the earth. Does Cleetus have a good case? Solution: We begin by drawing a free body diagram of the fish. For objects that are stationary relative to the surface of the earth, Newton’s second law gives This gives The true weight is mg, while the measured weight is Fscale. Therefore, the error is For objects at the surface of the earth, r=6.38x106m and . Error=0.34%. This is less than the 2% that the fish is underweight. Cleetus does not have a good case. Author: Kennedy, T. C. Subject: Dynamics Topic: Kinetics of Particles with Polar Coordinates Problem Length: 10 minutes Grading: Hand graded Presentation Format: Text and figure with problem statement Other: