More Problems

Problem 1:

Three particles are connected by rigid rods of negligible mass lying along the y-axis as shown.

If the system rotates about the x-axis with angular speed of 2 rad/s, find
(a) the moment of inertia about the x-axis and the total rotational kinetic energy evaluated from (1/2)Iw², and
(b) the linear speed of each particle and the total kinetic energy evaluated from å(1/2)m_iv_i².

Solution:

	Concepts, principles, relations that apply to the problem: The moment of inertia of a system about an axis
	Why do they apply? We are asked to find the moment of inertia and then use it in a simple calculation.
	How do they apply? (a) The moment of inertia is I = åm_ir_i². Here r_i is the perpendicular distance of particle i from the x-axis. I = 4 kg 9 m²+ 2 kg 4 m²+ 3 kg 16 m²= 92 kgm². The rotational kinetic energy is T_rot = (1/2)Iw²= 46´4/s²= 184 J. (b) The linear speed of particle i is v_i= wr_i. The linear speed of the 4 kg mass is v = 6 m/s, and its kinetic energy is (1/2)mv²= 72 J. The linear speed of the 2 kg mass is v = 4 m/s, and its kinetic energy is (1/2)mv²= 16 J. The linear speed of the 3 kg mass is v = 8 m/s, and its kinetic energy is (1/2)mv²= 96 . The sum of the kinetic energies of the three particles is 184 J.
	Details of the calculation: None

Problem 2:

The four particles in the figure below are connected by rigid rods.

The origin is at the center of the rectangle. If the system rotates in the x-y plane about the z-axis with an angular speed of 6 rad/s, calculate

	(a) the moment of inertia of the system about the z-axis and
	(b) the rotational energy of the system.

Solution:

	Concepts, principles, relations that apply to the problem: The moment of inertia about an axis, I = Sm_ir_i², the rotational kinetic energy, T_rot = (1/2)Iw²
	Why do they apply? We are asked to find he moment of inertia about and the rotational kinetic energy.
	How do they apply? (a) The moment of inertia is I = Sm_ir_i². Here r_i is the perpendicular distance of particle i from the z-axis. Each particle is a distance r = (9 + 4)^1/2m = Ö(13) m from the axis of rotation. I = (3 kg + 2 kg + 4 kg + 2 kg)(13 m²) = 1 43 kgm². (b) The rotational kinetic energy is T_rot = (1/2)Iw²= 71.5*36/s²= 2574 J.
	Details of the calculation: None

Problem 3:

Show that the moment of inertia with respect to the axis c of a solid uniform ellipsoid
(x²/a²) + (y²/b²) + (z²/c²) = 1 of mass M is given by M(a²+ b²)/5.

Solution:

	Concepts, principles, relations that apply to the problem: The moment of inertia
	Why do they apply? We are asked to find the moment of inertia with respect to an axis.
	How do they apply? The principal axes of the ellipsoid are the x-, y-, and z-axis. The moment of inertia about the z-axis is .
	Details of the calculation: Change variables to x' = x/a, y' = y/b, z' = z/c. Then . x'² + y'² + z'² = 1, we are integrating over the volume of a sphere. Change variables again to spherical coordinates. x' = r sinq cosf, y' = r sinq sinf, z' = r cosq. For an ellipsoid we have r = M/[(4p/3)abc]. Therefore I = (M/5)(a² + b²).

Problem 4:

A solid sphere toy globe of mass M and radius R rotates freely without friction with an initial angular velocity w₀. 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

Image417.gif (1325 bytes)

Useful integral:

Solution:

Concepts, principles, relations that apply to the problem:
Conservation of a component of the angular momentum

Why do they apply?
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.

How do they apply?

L_z = Iw = I₀w₀ = constant.
I₀ = (2/5)MR², I = I₀ + mR²sin²f.
I(t)w(t) = [(2/5)MR²+ mR²sin²f](dq/dt) = (2/5)MR²w₀.
constant speed: f = vt/R

Details of the calculation:
(dq/dt) = (2/5)MR²w₀/[(2/5)MR²+ mR²sin²(vt/R)]

Change variables: x = 2vt/R, sin²(x/2) = (1/2)(1-cosx), since òdx/(a +bcosx) is an integral found in integral tables. Use vT/R = p.

As m ® 0, Dq ® pw₀R/v = w₀T, which corresponds to the free rotation of the globe with angular speed w₀.

Problem 5:

A dumbbell consists of two spheres A and B, each with volume V, which are connected by a rigid rod. A has mass M and B has mass 2M. The distance between the centers of the spheres is d as shown below. In all parts of this problem assume that the mass and volume of the rod and the moment of inertia of each sphere about its diameter are so small that they can be taken to be zero, and that air resistance can be neglected.

	(a) True or false? If the dumbbell is dropped in a vacuum with the rod initially horizontal, the heavier sphere B will hit the floor first.
	(b) True or false? If the dumbbell is thrown on a frictionless horizontal surface with the rod horizontal, sphere B will move in a straight line with A rotating about it.
	(c) What is the moment of inertia of the dumbbell about an axis, which passes through the center of mass of the dumbbell and is perpendicular to the rod?
	(d) The dumbbell is placed on a frictionless horizontal table. Sphere A is attached to a frictionless pivot so that B can be made to rotate about A with constant angular velocity. If B makes one revolution in period T, what is the tension in the rod?
	(e) Sphere A is now attached to a frictionless pivot so that B hangs freely vertically. At some instant of time a strong wind begins to apply a constant horizontal force to B. As a result, the dumbbell rotates about A in a vertical plane. What is the speed of B (in terms of F, d, g, and M) at the instant when the dumbbell is horizontal?
	(f) The dumbbell is placed in water, which has density r. It is observed that by attaching a mass m to the rod, a distance l from the center of B, the dumbbell floats with the rod horizontal on the surface of the water and each sphere exactly half submerged, as shown below. The volume of the mass m is negligible. In terms of V, M, d, and r, derive the value of m and the value of l.

Solution:

Concepts, principles, relations that apply to the problem:
Various concepts and definitions introduced in a general physics course, such as Newton's second law, torque, buoyant force, moment of inertia, etc.

Why do they apply?
The different parts of the problem test the student's understanding of some of the fundamental concepts introduced in a typical general physics course.

How do they apply?

	(a) False: The only force acting on the dumbbell is gravity, giving mass-independent acceleration g.
	(b) False: The CM will move in a straight line. Both spheres may rotate about the CM with angular frequency w₀, if w₀ ¹ 0.
	(c) Location of the CM: MR = 2M(d-R), R = distance of CM from sphere A R = (2/3)d. moment of inertia: I = M(2d/3)² + 2M(d/3)² = (2/3)Md².
	(d) The tension must provide the centripetal force, F_c = 2Mv²/d. v = 2pd/T, F_c = 8Mp²d/T².
	(e) The tangential component of the applied horizontal force is F_t = Fcosq. The tangential component of the gravitational force is 2Mgsinq. The magnitude of the torque pointing out of the page is t = Fdcosq - 2Mgdsinq = Ia. (I = moment of inertia, I = 2Md², a = angular acceleration.) (1/2)Iw² = òtdq. w² = 2(Fd - 2Mgd)/I = (Fd - 2Mgd)/(Md²). But w² = v²/d². v = [(Fd - 2Mgd)/M]^1/2. We can a also use the work-kinetic energy theorem. Total work done by external forces: W = Fd - 2Mgd = (1/2)2Mv².
	(f) The system is at rest, the total force and the total torque are zero. No force: (1/2)rVg + (1/2)rVg - Mg - 2Mg -mg = 0, m = rV - 3M. (buoyant force + gravitational force = 0) No torque about sphere B: ((1/2)rVg - Mg )d - mgl = 0. l = ((1/2)rVd - Md)/(rV - 3M).

Details of the calculation:
None

Problem 6:

A 1500 kg automobile has a wheelbase (the distance between the axles) of 3 m. The center of mass of the automobile is on the centerline at a point 1.2 m behind the front axle. Find the force exerted by the ground on each wheel.

Solution:

	Concepts, principles, relations that apply to the problem: Static equilibrium: F = 0, t = 0
	Why do they apply? The car is in static equilibrium, F_tot= 0, t_tot= 0. The force of gravity Mg acts on the center of mass of the object. It produces no torque about the CM.
	How do they apply? F_tot= F₁+ F₂- Mg = 0. t_tot= F₂1.8 m - F₁1.2 m = 0. F₁= 1.5 F₂. 2.5F₂= Mg = 14700 N, F₂= 5880 N. F₁= 8820 N. The force exerted by the ground on each rear wheel is F₂/2 = 2940 N and the force exerted by the ground on each front wheel is F₁/2 = 4410 N.
	Details of the calculation: None

	Concepts, principles, relations that apply to the problem: Conservation of a component of the angular momentum
	Why do they apply? 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.
	How do they apply? L_z = Iw = I₀w₀ = constant. I₀ = (2/5)MR², I = I₀ + mR²sin²f. I(t)w(t) = [(2/5)MR²+ mR²sin²f](dq/dt) = (2/5)MR²w₀. constant speed: f = vt/R
	Details of the calculation: (dq/dt) = (2/5)MR²w₀/[(2/5)MR²+ mR²sin²(vt/R)] Change variables: x = 2vt/R, sin²(x/2) = (1/2)(1-cosx), since òdx/(a +bcosx) is an integral found in integral tables. Use vT/R = p. As m ® 0, Dq ® pw₀R/v = w₀T, which corresponds to the free rotation of the globe with angular speed w₀.