Assignment 3, solutions
Problem 1:
A cylindrical shell rolls without slipping down an incline as shown below. If it starts from rest, how far must it roll along the incline to obtain a speed v?
Solution:
Concepts: Energy conservation, rolling | |
Reasoning: Gravity is a conservative force. Rolling: ω = v/r. Energy conservation: ½mv^{2} + ½Iω^{2} = mgΔh. | |
Details of the calculation: I = mr^{2}, therefore v^{2} = gΔh. Δh = Δs*sinθ, therefore Δs = v^{2}/(g sinθ) is the distance the cylindrical shell must roll. |
Problem 2:
An inclined plane of length L makes an angle θ > 0^{o} with respect
to the horizontal. A very thin ring with radius R rolls down this plane. It
reaches the bottom in time Δt_{ring}.
(a) If a uniform density disk or sphere with the same radius R roll down this
plane, at what time do they reach the bottom? Find Δt_{disk}/Δt_{ring}
and Δt_{sphere}/Δt_{ring}.
(b) If all three object should reach the bottom at the same time Δt_{ring},
one needs to vary the angle θ for the disk and the sphere. Find sinθ_{disk}/sinθ
and sinθ_{sphere}/sinθ.
Solution:
Concepts: Rigid-body motion, energy conservation, kinematics | |
Reasoning: Energy conservation: mgLsinθ = ½mv^{2} + ½I(v^{2}/r^{2}) Kinematics: v^{2 }= 2aL, a = ½gsinθ, L = ½at^{2}, t = (2L/a)^{½} | |
Details of the calculation: (a) Ring: Energy conservation: mgLsinθ = ½mv^{2} + ½mr^{2}(v^{2}/r^{2}), gLsinθ = v^{2}. Kinematics: v^{2 }= 2aL, a = ½gsinθ, L = ½at^{2}, t = (2L/a)^{½} = (4L/(gsinθ))^{½}. Disk: Energy conservation: mgLsinθ = ½mv^{2} + ½(½mr^{2})(v^{2}/r^{2}), gLsinθ = ¾v^{2}. Kinematics: v^{2 }= 2aL, a = (2/3)gsinθ, L = ½at^{2}, t = (2L/a)^{½} = (3L/(gsinθ))^{½}. Sphere: Energy conservation: mgLsinθ = ½mv^{2} + ½(2/5)mr^{2}(v^{2}/r^{2}), gLsinθ = (7/10)v^{2}. Kinematics: v^{2 }= 2aL, a = (5/7)gsinθ, L = ½at^{2}, t = (2L/a)^{½} = (14/(5gsinθ))^{½}. Δt_{disk}/Δt_{ring} = (3/4)^{½}. Δt_{sphere}/Δt_{ring} = (7/10)^{½}. (b) The same acceleration will produce the same time Δt. Disk: (2/3)gsinθ_{disk} = ½gsinθ, sinθ_{disk}/sinθ = ¾. Sphere: (5/7)gsinθ_{disk} = ½gsinθ, sinθ_{disk}/sinθ = 7/10. |
Problem 3:
A rope rests on two platforms which are both inclined at angle θ (which you are free to pick) as shown. The rope has uniform density and its coefficient of friction with the platform is 1. The system has left-right symmetry. What is the largest possible fraction of the rope that does not touch the platform? What angle θ allows this maximum value?
Solution:
Concepts: Static equilibrium | |
Reasoning: Let the total mass of the rope be m and let a fraction f of it hang in the air. For each half of the symmetrical situation, the weight of the hanging fraction must be balanced by the vertical component of the tension in the rope. For the fraction of the rope touching the platform the frictional force must balance the gravitational force component along the plane plus the tension in the lower end. For a given angle θ: maximum frictional force --> maximum tension --> maximum hanging fraction | |
Details of the calculation: For the right half: Hanging weight = (f/2)mg. Vertical component of tension = T sinθ. (f/2)mg = T sinθ. T = (f/2)mg/sinθ. Mass of rope touching platform = (1 - f)m/2. Normal force = (1 - f)(m/2)g cosθ. Maximum frictional force = (1 - f)(m/2)g cosθ, since the coefficient of friction is 1. Gravitational force component along the plane = (1 - f)(m/2)g sinθ. (1 - f)(m/2)g cosθ = (1 - f)(m/2)g sinθ + (f/2)mg/sinθ. f = (cosθ sinθ - sin^{2}θ)/(1 + cosθ sinθ - sin^{2}θ). This can also be written as f = (sin2θ + cos2θ - 1)/(2 + sin2θ + cos2θ - 1). To find the angle for which f_{max}(θ) = f_{max}, set df/dθ = 0. θ_{max} = 22.5^{o}. f_{max} = (2^{1/2} - 1)/(2^{1/2} + 1) = 0.172. |
Problem 4:
A uniform disk of radius R and mass M is spinning about its diameter with angular velocity ω, as shown below. Located on the rim of the disk, at an angle θ from the spin axis, is point P, and P is moving with speed v_{p}. Point P is now suddenly fixed. Show that the subsequent linear speed v_{c} of the center of the disk is v_{c }= v_{p}/5.
Solution:
Concepts: Newton's second law: ΔP = FΔt, ΔL = τΔt | |
Reasoning: To reduce the velocity of point P to zero, an impulse must be delivered. The force acts at a point away from the CM. This results in a torque and an angular impulse about the CM. The impulse changes the momentum of the CM and the angular momentum about the CM in such a way that point P is at rest. This determines the magnitude of ΔP. The velocity of the CM is given by v_{c} = ΔP/M. | |
Details of the calculation: The disk rotates about the z-axis. Assume that when it is aligned with the yz-plane the impulse is delivered to the point P on the rim, a distance y = Rsinθ from the z-axis and z = Rcosθ from the y-axis. ΔP = FΔt i, ΔL = r × ΔP = -F R sinθ Δt k + F R cosθ Δt j. Just after the impulse, the momentum of the CM is ΔP and the angular momentum about the CM is L = I_{z}ωk - F R sinθ Δtk + F R cosθ Δtj. The velocity of the point P after the impulse is v = (FΔt/M)i + ω × r = 0. (FΔt/M)i - (1/I_{z})(I_{z}ω - F R sinθ Δt)R sinθ i + (1/I_{y})(F R cosθ Δt)R cosθ i = 0. For a disk of radius R we have: I_{z} = I_{y} = 2∫_{0}^{R}∫_{0}^{π}rdrdφ(r sinφ)^{2}ρ = MR^{2}/4. Therefore: FΔt/M - ωR sinθ - FR^{2}sin^{2}θ Δt/(MR^{2}/4) + FR^{2}cos^{2}θ Δt/(MR^{2}/4) = 0. FΔt = (M/5)ωR sinθ . The speed of point P before the impulse is v_{p} = ωR sinθ. Therefore FΔt = (M/5)v_{p}, v_{c} = FΔt/M = v_{p}/5. |
Problem 5:
A solid sphere toy globe of mass M and radius R rotates freely without
friction with an initial angular velocity ω_{0}. A bug of mass m starts
at one pole N and travels with constant speed v to the other pole S along a
meridian in time T. The axis of rotation of the globe is held fixed. Show that
during the time the bug is traveling the globe rotates through an angle
Δθ = (πω_{0}R/v) (2M/(2M + 5m))^{1/2}.
Useful integral: ∫_{0}^{2π}dx/(a + b cosx) = 2π/(a^{2} - b^{2})^{1/2}, (a^{2} > b^{2}).
Solution:
Concepts: Conservation of a component of the angular momentum | |
Reasoning: The axis of rotation is held fixed, so external forces cannot be excluded. But holding an axis fixed cannot exert a torque about this axis. So the component of angular momentum about the axis of rotation is conserved. | |
Details of the calculation: L_{z} = Iω = I_{0}ω_{0} = constant. I_{0} = (2/5)MR^{2}, I = I_{0} + mR^{2}sin^{2}φ. I(t)ω(t) = [(2/5)MR^{2}_{ }+ mR^{2}sin^{2}φ](dθ/dt) = (2/5)MR^{2}ω_{0}. constant speed: φ = vt/R. (dθ/dt) = (2/5)MR^{2}ω_{0}/[(2/5)MR^{2}_{ }+ mR^{2}sin^{2}(vt/R)]. Δθ = ∫_{0}^{T}dt (2/5)MR^{2}ω_{0}/[(2/5)MR^{2}_{ }+ mR^{2}sin^{2}(vt/R)]. Change variables: x = 2vt/R, sin^{2}(x/2) = ½(1 - cosx), since ∫dx/(a + b cosx) is an integral found in integral tables and is given here.. Use vT/R = π. Δθ = (Rω_{0}/v)∫_{0}^{2π}dx/[2_{ }+ (5m/M) sin^{2}(x/2)] = (Rω_{0}/v)∫_{0}^{2π}dx/[2_{ }+ (5m/(2M))(1 - cosx)] = (πRω_{0}/v)(2M/(2M + 5m))^{1/2}. As m --> 0, Δθ --> πω_{0}R/v = ω_{0}T, which corresponds to the free rotation of the globe with angular speed ω_{0}. |