As shown in Figure 10.13 on page 306, toppling chimneys often break apart in midfall because the mortar between the bricks cannot withstand much shear stress. As the chimney begins to fall, shear forces must act on the topmost sections to accelerate them tangentially so that they can keep up with the rotation of the lower part of the stack. For simplicity, let us model the chimney as a uniform rod of length ℓ pivoted at the lower end. The rod starts at rest in a vertical position (with the frictionless pivot at the bottom) and falls over under the influence of gravity. What fraction of the length of the rod has a tangential acceleration greater than g sin θ, where θ is the angle the chimney makes with the vertical axis?
As shown in Figure 10.13 on page 306, toppling chimneys often break apart in midfall because the mortar between the bricks cannot withstand much shear stress. As the chimney begins to fall, shear forces must act on the topmost sections to accelerate them tangentially so that they can keep up with the rotation of the lower part of the stack. For simplicity, let us model the chimney as a uniform rod of length ℓ pivoted at the lower end. The rod starts at rest in a vertical position (with the frictionless pivot at the bottom) and falls over under the influence of gravity. What fraction of the length of the rod has a tangential acceleration greater than g sin θ, where θ is the angle the chimney makes with the vertical axis?
As shown in Figure 10.13 on page 306, toppling chimneys often break apart in midfall because the mortar between the bricks cannot withstand much shear stress. As the chimney begins to fall, shear forces must act on the topmost sections to accelerate them tangentially so that they can keep up with the rotation of the lower part of the stack. For simplicity, let us model the chimney as a uniform rod of length ℓ pivoted at the lower end. The rod starts at rest in a vertical position (with the frictionless pivot at the bottom) and falls over under the influence of gravity. What fraction of the length of the rod has a tangential acceleration greater than g sin θ, where θ is the angle the chimney makes with the vertical axis?
A skateboarder with his board can be modeled as a particle of mass 80.0 kg, located at his center of mass. As shown in the figure below, the skateboarder starts from rest in a crouching position at one lip of a half-pipe (point). On his descent, the skateboarder moves without friction so
that his center of mass moves through one quarter of a circle of radius 6.20 m.
i
(a) Find his speed at the bottom of the half-pipe (point Ⓡ).
m/s
(b) Immediately after passing point Ⓑ, he stands up and raises his arms, lifting his center of mass and essentially "pumping" energy into the system. Next, the skateboarder glides upward with his center of mass moving in a quarter circle of radius 5.71 m, reaching point D. As he
passes through point ①, the speed of the skateboarder is 5.37 m/s. How much chemical potential energy in the body of the skateboarder was converted to mechanical energy when he stood up at point Ⓑ?
]
(c) How high above point ① does he rise?
m
A 31.0-kg child on a 3.00-m-long swing is released from rest when the ropes of the swing make an angle of 29.0° with the vertical.
(a) Neglecting friction, find the child's speed at the lowest position.
m/s
(b) If the actual speed of the child at the lowest position is 2.40 m/s, what is the mechanical energy lost due to friction?
]
Chapter 10 Solutions
Physics for Scientists and Engineers, Technology Update, Hybrid Edition (with Enhanced WebAssign Multi-Term LOE Printed Access Card for Physics)
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