Concept explainers
(a)
The speed at the bottom of the half-pipe.
(a)
Answer to Problem 11.48AP
The speed at the bottom of the half-pipe is
Explanation of Solution
Given info: The mass of particle is
Write the expression for conservation of work and energy law.
Here,
The skateboarder is at rest at point A, so there is a potential energy at point A,
Here,
The center of mass moves through one quarter of the circle.
The radius of the circle is,
The skateboarder is in motion so it acquires the kinetic energy at point B,
Here,
Substitute
Substitute
Conclusion:
Therefore, the speed at the bottom of the half-pipe is
(b)
The
(b)
Answer to Problem 11.48AP
The angular momentum of him about the center of curvature at the point B is
Explanation of Solution
Given info: The mass of particle is
Write the expression for the angular momentum about the center of curvature.
Here,
Substitute
Conclusion:
Therefore, the angular momentum of him about the center of curvature at the point B is
(c)
To explain: The angular momentum of him is constant in this maneuver, whereas the kinetic energy of his body is not constant.
(c)
Answer to Problem 11.48AP
After the passing point B, there is no torque about the axis of the channel act on him so; the angular momentum will be constant, but his legs convert the chemical energy into mechanical energy and the kinetic energy of his body is not constant.
Explanation of Solution
Given info: The mass of particle is
A skateboarder passes the point B, so there is no tangential force acts on him because the wheels on the skate prevent this force. The torque is zero due to no tangential force, so the angular momentum will be constant.
The kinetic energy increase because his legs convert chemical energy into mechanical energy and the kinetic energy will not be constant. While the normal force rises trajectory to enhance his linear momentum.
Conclusion:
Therefore, after the passing point B, there is no torque about the axis of the channel act on him so; the angular momentum will be constant, but his legs convert the chemical energy into mechanical energy and the kinetic energy of his body is not constant.
(d)
The speed immediately after the skateboarder stands up.
(d)
Answer to Problem 11.48AP
The speed of skateboarder after he stands up is
Explanation of Solution
Given info: The mass of particle is
The skateboarder stands up, so the distance is,
Write the expression for angular momentum.
Here,
Substitute
Conclusion:
Therefore, the speed of skateboarder after he stands up is
(e)
The amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up.
(e)
Answer to Problem 11.48AP
The amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up is
Explanation of Solution
Given info: The mass of particle is
At point B, the skate boarder has kinetic and chemical energy is,
Here,
At point C, he has kinetic energy due and the potential energy is,
Here,
Write the expression of the conservation of energy.
Substitute
Write the expression for the kinetic energy at point B.
Substitute
Thus, the kinetic energy at point B is
Write the expression for the kinetic energy at point C.
Substitute
Thus, the kinetic energy at point C is
Write the expression for potential energy at point C.
Here,
The radius of the pipe at point C,
Substitute
Thus, the potential energy at point C is
Substitute
Conclusion:
Therefore, the amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up is
Want to see more full solutions like this?
Chapter 11 Solutions
PHYSICS 1250 PACKAGE >CI<
- You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are present for the testing of the ride, in which an empty 190 kg car is sent along the entire ride. Near the end of the ride, the car is at near rest at the top of a 102 m tall track. It then enters a final section, rolling down an undulating hill to ground level. The total length of track for this final section from the top to the ground is 250 m. For the first 230 m, a constant friction force of 400 N acts from computer-controlled brakes. For the last 20 m, which is horizontal at ground level, the computer increases the friction force to a value required for the speed to be reduced to zero just as the car arrives at the point on the track at which the passengers exit. (a) Determine the required constant friction force (in N) for the last 20 m for the empty test car. 4905.89 N (b) Find the highest speed (in m/s) reached by the car during the final section of track…arrow_forwardThree dimensions. Three point particles are fixed in place in an xyz coordinate system. Particle A, at the origin, has mass ma. Particle B, at xyz coordinates (3.00d, 1.00d, 3.00d), has mass 4.00ma, and particle C, at coordinates (-2.00d, 1.00d, -1.00d), has mass 2.00ma. A fourth particle D, with mass 3.00ma, is to be placed near the other particles. If distance d = 4.50 m, at what (a) x, (b) y, and (c) z coordinate should D be placed so that the net gravitational force on A from B, C, and D is zero? (a) Number Units (b) Number |Units (c) Number Unitsarrow_forwardAn object with mass M slides across a frictionless half-circular slide of radius R. The object is initially held stationary at the top of the slide. The initial height is, hi = R. The mass is released and slides down, around the bottom, and back up the other side. The radius of the slide is R = 5.79 m and the mass of the object is M = 37 kg. When the object is moving up the slide, at an angle θ = 18o from the vertical, what is the normal force on the object?arrow_forward
- Jane is sitting on a chair with her lower leg at a 30.0° angle with respect to the vertical, as shown. You need to develop a computer model of her leg to assist in some medical research. Assume that her leg can be modeled as two uniform cylinders, one with mass M = 22.0 kg and length L = 35.0 cm and one with mass m = 10.0 kg and length l = 40.0 cm. Find the x-component of the center of mass of Jane’s leg.arrow_forwardThe H2S molecule contains two hydrogen atoms and one sulfur atom. The position vectors of the hydrogen atoms and the sulfur atom are rHI=(0.70, 1.01, 0.00), rí2=(0.70, -1.01, 0.00), rs=(0.00, 0.00, 0.00), respectively. Providing that the mass of hydrogen atom is 1 amu and the mass of sulfur atom is 32 amu. Find a. The angle between the two H-S bonds. b. The position vector of the centre of mass of the H2S molecule. c. The position vectors of the hydrogen atoms and the sulfur atom with respect to the centre of the mass of the molecule (pointing from the centre of the mass of the molecule to the hydrogen atoms and the sulfur atom).arrow_forwardJane is sitting on a chair with her lower leg at a 30.0° angle with respect to the vertical, as shown. You need to develop a computer model of her leg to assist in some medical research. Assume that her leg can be modeled as two uniform cylinders, one with mass M = 22.0 kg and length L = 35.0 cm and one with mass m = 10.0 kg and length l = 40.0 cm. Find the y-component of the center of mass of Jane’s leg.arrow_forward
- Three solid, uniform boxes are aligned as in the figure below. Find the x- and y-coordinates (in m) of the center of mass of the three boxes, measured from the bottom left corner of box A. (Consider the three-box system.) HINT X cm y em || = Origin E E m 0.360 m A 0.750 kg 0.420 m B 1.00 kg 0.240 m с 0.650 kg 0.410 m Xarrow_forwardThree solid, uniform boxes are aligned as in the figure below. Find the x- and y-coordinates (in m) of the center of mass of the three boxes, measured from the bottom left corner of box A. (Consider the three-box system.) HINT Ycm Origin 0.200 m 0.850 kg 0.280 m B 1.00 kg 0.160 m C 0.650 kg 0.362 marrow_forwardA 3.0-kg mass is sliding on a horizontal frictionless surface with a speed of V=3.0 m/s when it collides with a 1.0-kg mass initially at rest as shown in the figure. The masses stick together and slide up a frictionless circular track of radius 0.40 m, as the drawing below shows. To what maximum height, h, above the horizontal surface will the masses slide. 7. 040 marrow_forward
- Find a unit vector which is in the xy plane (no k-component) which is also perpendicular to the force 20i + 30j + 60k N. Please do not use cross product.arrow_forwardIn the instant of the figure, two particles move in an xy plane. Particle P₁ has mass 7.40 kg and speed v₁ = 1.91 m/s, and it is at distance d₁ = 9.52 m from point O (the figure is not drawn to scale). Particle P₂ has mass 2.73 kg and speed v2 = 2.37 m/s, and it is at distance d₂ = 8.89 m from point O. What is the magnitude of the net angular momentum of the two particles about O? P₁O Number i Units ས། ྋ P2arrow_forwardThe coefficient of friction between the block of mass m1 = 3.00 kg and the surface in the figure uk = 0.480. The system starts from rest. What is the speed of the ball of mass m2 5.00 kg when it has fallen a distance h = 1.90 marrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning