Physics 11 Fall 2012 Practice Problems 6 - Solutions

Transcription

1 Physics 11 Fall 01 Practice Problems 6 - s 1. Two points are on a disk that is turning about a fixed axis perpendicular to the disk and through its center at increasing angular velocity. One point is on the rim and the other point is halfway between the rim and the center. (a) Which point moves the greater distance in a given time? (b) Which point turns through the greater angle? (c) Which point has the greater speed? (d) Which point has the greater angular speed? (e) Which point has the greater tangential acceleration? (f) Which point has the greater angular acceleration? (g) Which point has the greater centripetal acceleration? (a) The point on the rim travels a greater distance in the same amount of time, since it has to make a bigger orbit. (b) Both points move through the same angle. (c) The speed is v = rω. Since point particles change their angle by the same amount in the same time, the point with the bigger radius has the bigger velocity. So, the point on the rim moves with the greater speed. (d) Since the angular speed is ω, and both points travel through the same angle in the same time, they both have the same angular speed. (e) The problem states that the disk is turing at increasing angular velocity, and so ω is increasing. Thus, v = rω is increasing. The tangential acceleration is a tang = rα. So, the particle with the bigger radius has the bigger tangential acceleration; i.e., the point on the rim has the bigger tangential acceleration. (f) The angular acceleration is α, and is the same for both points. (g) The centripetal acceleration is a cent = v /r = (rω) /r = rω. Since ω is the same for both points, the point on the rim has the bigger centripetal acceleration, since it has the bigger value of r. 1

2 . What is the angular speed of Earth in radians per second as it rotates about its axis? The angular speed is ω = θ. The Earth rotates through an angle of π radians in 4 t hours. 4 hours is seconds, or 86,400 seconds. So, the angular speed is ω Earth = θ t = π = rad/sec.

3 I = [ ( kg)( m) + ( kg)( 0.00 m) ] + ( kg)( m) 5 m = kg The Thepercent methane difference molecule between (CH 4 ) I has and four Iapp is: hydrogen atoms located at the vertices of a regular tetrahedron of edge length 0.18 I Iapp kg m kg m nm, with the carbon atom = at the center = 3.6 % of the tetrahedron. I Find the moment kg of m inertia of this molecule for rotation about an axis that passes through the centers of (b) The rotational inertia would increase because I the carbon atom and one of the hydrogen cm of a hollow sphere is greater than atoms. I cm of a solid sphere. 49 The methane molecule (CH 4 ) has four hydrogen atoms located at the vertices of a regular tetrahedron of edge length 0.18 nm, with the carbon atom at the center of the tetrahedron (Figure 9-48). Find the moment of inertia of this molecule for rotation about an axis that passes through the centers of the carbon atom Because and the one axis of the of rotation hydrogen is through atoms. the carbon and one of the hydrogens, neither of these atoms contribute. The moment of Picture inertia isthe then Problem due only tothe the remaining axis of H rotation three hydrogens. passes through The moment the center of inertia of the is base of the tetrahedron. The carbon 3 3 atom and I the = hydrogen m i ri = matom H at ri the, apex of the tetrahedron i=1 do not contribute i=1 to I because since all the of thedistance masses are of identical. their nuclei Now from the radii the axis are also of rotation all equal is iszero. just From the C H the distance geometry, from the central distance hydrogen of the to three the atoms on the vertex. if the distance between the hydrogens is a, then the radius H nuclei from the rotation axis is a is/ 3, where a is the length of a side r = a 1 cos 30 = a = a H of the tetrahedron.. a 3 3 Apply the definition of the moment I = miri = mhr1 + mhr + mhr So, we find that i of inertia for a system of particles to ( ) obtain: I = 3m H a 3 a = m H = a mh = mha 3 = ( ) ( ) = kg m. Substitute numerical values and 7 I = kg m evaluate I: = kg m 3 ( )( ) H 3

4 4. During most of its lifetime, a star maintains an equilibrium size in which the inward force of gravity on each atom is balanced by an outward pressure force due to the heat of the nuclear reactions in core. But after all the hydrogen fuel is consumed by nuclear fusion, the pressure force drops and the star undergoes gravitational collapse until it becomes a neutron star. In a neutron star, the electrons and protons are squeezed together by gravity until they form neutrons. Neutron stars spin very rapidly and emit intense pulses of radio and light waves, one pulse per rotation. These pulsing stars were discovered in the 1960s and are called pulsars. (a) A star with the mass (M = kg) and size (R = m) of our sun rotates once every 30 days. After undergoing gravitational collapse, the star forms a pulsar that is observed by astronomers to emit radio pulses every 0.10 seconds. By treating the neutron star as a solid sphere, deduce its radius. (b) What is the speed of a point on the equator of the neutron star? Your answer will be somewhat too large because a star cannot be accurately modeled as a solid sphere. Even so, you will be able to show that a star, whose mass is 10 6 larger than the earth s, can be compressed by gravitational forces to a size smaller than a typical state in the United States! (a) As the star collapses, it speeds up by the conservation of angular momentum. It s initial angular momentum is L i = Mv i R i = MRi ω i = πf i MRi, where we have recalled that v = Rω, and that ω = πf, where f is the rotation frequency. The angular momentum after the collapse is, then, L f = πf f MRf. Conservation of angular momentum says that L i = L f, and so R f = f i Tf R i = R i, f f T i where we ve re-expressed the radius in terms of the rotation period. Now, the initial frequency is once per 30 days, so T i = = seconds, while T f = 0.10 seconds. Thus, R f = Tf 0.1 R i = T i = m = 137 km. (b) The speed of a point on the equator is v = R f ω f = πr f /T = π( )/0.1 = m/s. 4

5 us sign is a consequence of the fact that the frictional torque otion of the wheel. 5. A pendulum consisting of a string of length L attached to a bob of mass m swings in a vertical plane. When the string is at an angle θ to the vertical, endulum consisting of a string of length L attached to a bob of (a) calculate the tangential acceleration of the bob using F t = ma t. s in a vertical (b) What plane. is the torque When exerted the about string the pivot point? is at an angle θ to the vertic (c) Show that τ = Iα with a t = Lα gives the same tangential acceleration as found in Part (a). t = ma t, calculate the tangential acceleration of the bob? (b) Wh xerted about the pivot point? (c) Show that τ = I α with a t = L tangential acceleration as found in Part (a). (a) The pendulum is seen in the figure to the right. The only tangential force is the component of gravity along the rotation path, Problem The pendulum acting on it are shown in diagram. Note that the F t = mg sin θ. Since F t = ma t, then we just solve for the acceleration, a t = F t = g sin θ. string is radial, and m so ential force on the ball. ewton s nd law in both nd rotational form to find l component of the f the bob. (b) The torque is just τ = rf t, where r = L is the distance from the pivot point. So, τ = mgl sin θ. (c) The torque is τ = Iα. Since the string is massless and all the mass is in the bob, then I = mr = ml. So, τ = mgl sin θ = ml α. Furthermore, since α = τ/i = a t /L, then a t = τl, or I a t = τl I o the free-body ess the component of ent to the circular path = mgl sin θ ml = g sin θ, which is the same result we found in part (a). F t = mg sinθ L θ T r mg r θ nd law to express the 5 Ft mg sinθ a = = = g sinθ

6 6. A uniform 1.5 m diameter ring is pivoted at a point on its perimeter so that it is free to rotate about a horizontal axis that is perpendicular to the plane of the ring. The ring is released with the center of the ring at the same height as the axis. (a) If the ring was released from rest, what was its maximum angular speed? (b) What minimum angular speed must it be given at release if it is to rotate a full 360? (a) When the ring is just hanging straight down, it has zero potential energy. When it s released from it s initial height, h = R, then it has potential energy mgh = mgr. Now, at the bottom of the path, all the energy is kinetic. Since it s rotating, the energy is all rotational kinetic energy, KE = 1 Iω. What s I? We can look up the moment of inertia of the ring about an axis through the center to find I = mr. Since the ring is rotating about it s rim, then we can use the parallel axis theorem to write the moment of inertia about the rim as I rim = I center +md = mr + mr = mr. So, the kinetic energy is KE = 1 Iω = mr ω. So, if energy is conserved, then KE = P E mr ω = mgr. Solving for ω gives g ω = R. Plugging in the numbers gives ω = g R = = 3.61 rad/sec. (b) Now we want to figure out how fast we d need to rotate it around in order for it to rotate to the top. In this case, it still has the same kinetic energy, for some ω, KE = mr ω. Now it has to get to the top, changing its height by R, so that it s final potential energy is P E = mgr. Solving for KE = P E ω = g/r, exactly the same as in part (a). 6

7 A uniform solid sphere of mass M and radius R is free rotate about a horizontal axis through its center. A string is wrapped around the sphere and is attached to an object of mass m. Assume that the string does not slip on the sphere. Find (a) the acceleration of the object and (b) the tension in the string. Chapter 9 77 A uniform solid sphere of mass M and radius R is free t horizontal axis through its center. A string is wrapped around the s attached to an object of mass m (Figure 9-58). Assume that the stri on the sphere. Find (a) the acceleration of the object, and (b) the te string. Picture the Problem The force diagram shows the forces acting and the hanging object. The tension in the string is responsible acceleration of the sphere and the difference between the weight o the tension is the net force acting on the hanging object. We can us law to obtain two equations in a and T that we can solve simultane (a) We can draw the force diagram for the system as seen in to the right. The weight of the block creates a torque on the wheel, causing it to rotate. We can write down the net forces and torques acting on the system. For the sphere, τ = T R = Iα. While for the hanging mass, calling the vertical direction x, Fx = T mg = ma. 0 M (a) Noting that T = T, apply Newton s nd law to the sphere and the hanging object: Substitute for I sphere and α in equation (1) to obtain: Eliminate T between equations () and (3) and solve for a to obtain: F R x T x T mg τ 0 = TR = I sphereα and F x = mg T = ma Now, since the string isn t slipping around the wheel, the angular acceleration is just α = a/r. Furthermore, the moment of inertia for a sphere is I = 5 MR. So, we find the following system of equations T = Ma 5 T = m (g a). a = a ( MR ) R Setting these two expressions equal to each other and solving for the acceleration gives TR = 5 g a =. 1 + M 5m g M 1+ 5m (b) Now we just substitute this result back in to the expression T = Ma to find 5 T = 5 Mg 1 + M 5m = Mmg 5m + M. (b) Substitute for 7 a in equation () and solve for T to obtain: T = mmg 5m + M

8 8. A basketball rolls without slipping down an incline of angle θ. The coefficient of static friction is µ s. Model the ball as a thin spherical shell. Find Picture the Problem The three forces acting on the basketball a the ball, the normal force, and the force of friction. Because th assumed to be acting at the center of mass, and the normal force center of mass, the only force which exerts a torque about the c the frictional force. Let the mass of the basketball be m and app (a) the acceleration of the center of mass of the ball, (b) the frictional force acting on the ball, and (c) the maximum angle of the incline for which the ball will roll without slipping. law to find a system of simultaneous equations that we can solve f (a) We can begin called by writing for in down the problem Newton s statement. laws for the ball, including the torques: y Fx = mg sin θ + F f = ma Fy = mg cos θ + F n = 0 F r n τi = F f R = Iα. Furthermore, since the ball is rolling without slipping, we have that a = αr. So, we have enough to solve this system of equations. Plugging in F f = Iα/R = Ia/R to the first equation and solving for a gives a = mg sin θ m + I/R. x m r 0 mg r θ f r (a) Apply Newton s nd law in both translational and mg rotational sin θ form to a = the ball: m + m/3r = 3 g sin θ. 5 The moment of inertia of the spherical shell is just I = 3 mr, and so (b) Since F f = I a = R 3 mr a = ma, we have R 3 F f = mg sin θ. 5 Because the basketball is rolling without slipping we know that: Substitute in equation (3) to obtain: From Table 9-1 we have: F = mg f x sinθ s = F = cos y Fn mg θ = and τ 0 = f sr = I 0α a α = r (c) Now, we know that the frictional force is F f = µ s F N = µ s mg cos θ. Setting this equal to our result from part (b) we find 5 mg sin θ = µ smg cos θ tan θ = 5 µ s. f r I0 So, the maximum angle of the incline that the ball will roll without slipping is ). θ = tan 1 ( 5 µ s The rolling of the ball increases the angle from the sliding case. s = 3 a r I 0 = mr Substitute for I 0 and 8 α in equation (4) and solve for f s : a ( mr ) f fs r = 3 s = r

9 9. Released from rest at the same height, a thin spherical shell and a solid sphere of the same mass m and radius R roll without slipping down an incline through the same vertical drop H. Each is moving horizontally as it leaves the ramp. The spherical shell hits the ground a horizontal distance L from the end of the ramp and the solid sphere hits the ground a distance L from the end of the ramp. Find the ratio L /L. The distance that each ball goes after leaving the ramp depends on it s speed upon leaving the ramp. Since both balls take the same amount of time to reach the ground then L = v t, while L = v t, where v is the speed of the shell and v is the speed of the sphere at the bottom of the ramp. Now, the ratio L /L = v /v. So, we need to know the speed of the balls. Each ball starts off with the same potential energy, mgh. The final energy of each ball at the bottom of the ram is made up of translational and rotational kinetic energy, KE shell = 1 mv + 1 I shellω KE sphere = 1 mv + 1 I sphereω Now, since the balls are rolling without slipping, ω = v/r, while ω = v /R. Plugging in these expressions gives KE shell = ( I shell ) mr mv ( ) KE sphere = I sphere mv mr Since energy is conserved KE shell = KE sphere = mgh. So, mgh = ( ) I shell mr mv ( ) mgh = I sphere mv mr Taking the ratio v /v gives v v = 1 + I shell mr 1 + I sphere mr. Since I shell = 3 mr, while I sphere = 5 mr, we have v v = L L = 1 + / /5 = 1 =

10 10. According to the Standard Model of Particle Physics, electrons are pointlike particles having no spatial extent. (This assumption has been confirmed experimentally, and the radius of the electron has been shown to be less than meters.) The intrinsic spin of an electron could in principle be due to its rotation. Let us check to see if this conclusion is feasible. (a) Assuming that the electron is a uniform sphere whose radius is m, what angular speed would be necessary to produce the observed intrinsic angular momentum of /? (b) Using this value of the angular speed, show that the speed of a point on the equator of a spinning electron would be moving faster than the speed of light. What is your conclusion about the spin angular momentum being analogous to a spinning sphere with spatial extent? (a) The angular momentum can be expressed in terms of the moment of inertia, I, and the angular velocity, ω, as L = Iω. Thus, the angular speed is ω = L/I. Now, for a uniform sphere, I = 5 MR, and for L = /, we find ω = L I = Taking M = m e = kg, we have ω = 5 4MR = 5 4MR (10 18 ) = rad/sec. (b) The speed is v = Rω, so using our results from part (a) gives v = Rω = = m/s! This is much faster than light. So, if we require that these points on the equator can t spin faster than light, we have to abandon the model of the electron as a tiny little spinning sphere. 10