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Physics 11 Fall 2012 Practice Problems 6 - Solutions 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? ———————————————————————————————————— Solution (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 atang = 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 acent = v 2 /r = (rω)2 /r = rω 2 . 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? ———————————————————————————————————— Solution . The Earth rotates through an angle of 2π radians in 24 The angular speed is ω = ∆θ ∆t hours. 24 hours is 24 × 3600 seconds, or 86,400 seconds. So, the angular speed is ωEarth = ∆θ 2π = = 7.3 × 10−5 rad/sec. ∆t 86400 2 [ ] I = 2 52 (0.500 kg )(0.0500 m ) + (0.500 kg )(0.200 m ) + 121 (0.0600 kg )(0.300 m ) 2 2 2 = 0.0415 kg ⋅ m 2 difference andfour I app is: 3.The Thepercent methane moleculebetween (CH4 ) Ihas hydrogen atoms located at the vertices of a regular tetrahedron of edge length 0.18 I−I 0.0415 kg ⋅ m 2 − 0.0400 kg ⋅ m 2 nm, with the carbonappatom = at the center = 3.6 % 2 I 0.0415 of the tetrahedron. Find the moment of kg ⋅ m inertia of this molecule for rotation about an axis that passes through the centers of (bthe ) The rotational inertia would increase because Icm of a hollow sphere is greater carbon atom and one of the hydrogen than Icm of a solid sphere. atoms. 49———————————————————————————————————— •• The methane molecule (CH4) 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 Solution 9-48). Find the moment of inertia of this molecule for rotation about an axis that passes through the centers of the carbon Because of rotation is through atom and the oneaxis of the hydrogen atoms. the carbon and one of the hydrogens, neither of these atoms contribute. The moment of Picture axis of inertia is the then Problem due only toThe the remaining rotation passes through the center of theis three hydrogens. The moment of inertia base of the 3tetrahedron. The carbon 3 X X atom andI the atom atri2the = hydrogen mi ri2 = m , apex H i=1 i=1 of the tetrahedron do not contribute to I because of identical. their nuclei since all the of thedistance masses are Now the radii are also all equalis iszero. is just the from the axis of rotation From distance from the hydrogen the the geometry, the central distance of the to three atoms on the vertex. if the distance beH nuclei from the rotation axis is tween the hydrogens is a, then the radius ais/ 3 , where a is the length of a side a 2 a a 1 = √ =√ . r= of the tetrahedron. ◦ 2 cos 30 2 3 H C H H H a 3 Apply the definition of the moment So, we find that of inertia for a system of particles to I = obtain: I = ∑ mi ri 2 = mH r12 + mH r22 + mH r32 Substitute numerical values and evaluate I: I = 1.67 × 10 − 27 kg 0.18 × 10 −9 m i 2 2 3mH ⎛ a ⎞ ⎟ = mH a 2 = mH=a 3mH ⎜ ⎝ 3 ⎠2 = (1.67 × 10−27 ) (0.18 × 10−9 ) = 5.41 × 10−47 kg m2 . √a 3 2 ( )( = 5.4 × 10 − 47 kg ⋅ m 2 3 ) 2 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 = 2.0 × 1030 kg) and size (R = 7.0 × 108 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 106 larger than the earth’s, can be compressed by gravitational forces to a size smaller than a typical state in the United States! ———————————————————————————————————— Solution (a) As the star collapses, it speeds up by the conservation of angular momentum. It’s initial angular momentum is Li = M vi Ri = M Ri2 ωi = 2πfi M Ri2 , where we have recalled that v = Rω, and that ω = 2πf , where f is the rotation frequency. The angular momentum after the collapse is, then, Lf = 2πff M Rf2 . Conservation of angular momentum says that Li = Lf , and so s r fi Tf Rf = Ri = Ri , ff Ti where we’ve re-expressed the radius in terms of the rotation period. Now, the initial frequency is once per 30 days, so Ti = 3600 × 24 × 30 = 2.6 × 106 seconds, while Tf = 0.10 seconds. Thus, r r Tf 0.1 Ri = 7 × 108 = 1.37 × 105 m = 137 km. Rf = Ti 2.6 × 106 (b) The speed of a point on the equator is v = Rf ωf = 2πRf /T = 2π(1.37×105 )/0.1 = 8.6 × 106 m/s. 4 ⎝ ⎠ nus sign is a consequence of the fact that the frictional torque motion 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, P (a) calculate the tangential acceleration of the bob using Ft = mat . pendulum consisting of a string of length L attached to a bob of (b) What is the torque exerted about the pivot point? gs in a vertical plane. P When the string is at an angle θ to the vertic (c) Show that τ = Iα with a = Lα gives the same tangential acceleration as found in Part (a).the tangential acceleration of the bob? (b) Wh Ft = ma t , calculate t exerted about the pivot point? (c) Show that ∑τ = Iα with a t = L ———————————————————————————————————— Solution The pendulum is seen the figure toin the Part (a). e tangential(a)acceleration asin found right. The only tangential force is the component of gravity along the rotation path, Problem The pendulum F = mg sin θ. s acting on it are Since Fshown = ma , thenin we just solve for the acceleration, y diagram. Note that the F = g sin θ. a = e string is radial, and mso (b) The torque is just τ = rF , where r = L is gential force on the ball. nd the distance from the pivot point. So, Newton’s 2 law in τ both = mgL sin θ. and rotational form to find (c) The torque is τ = Iα. Since the string is massless and allthe the mass is in the bob, al component of then I = mr = mL . So, τ = mgL sin θ = mL α. Furthermore, since α = τ /I = of the bob. t t θ t t t L t 2 r T 2 2 at /L, then at = at = τL , I rθ mg or τL mgL2 sin θ = = g sin θ, I mL2 which is the same result we found in part (a). to the free-body ress the component of gent to the circular path nd s 2 law to express the Ft = mg sin θ 5 a = Ft = mg sin θ = g sin θ 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◦ ? ———————————————————————————————————— Solution (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 = 12 Iω 2 . What’s I? We can look up the moment of inertia of the ring about an axis through the center to find I = mR2 . 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 Irim = Icenter +md2 = mR2 + mR2 = 2mR2 . So, the kinetic energy is KE = 12 Iω 2 = mR2 ω 2 . So, if energy is conserved, then KE = P E ⇒ mR2 ω 2 = mgR. Solving for ω gives r g . ω= R Plugging in the numbers gives r r g 9.8 ω= = = 3.61 rad/sec. R 0.75 (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 = mR2 ω 2 . Now it has to get to the top, changing its height by R, so p 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). 6 890 Chapter 9 7. A uniform solid sphere of mass M and radius R is free rotate about a horizontal axis •• is wrapped A uniform solid sphere of mass M and radius R is free t through its center. 77 A string around the sphere and is attached to through an horizontal axis its center. A string is wrapped around the s object of mass m. Assume that the string attached to an object of mass m (Figure 9-58). Assume that the stri does not slip on the sphere. Find (a) on the sphere. Find (a) the acceleration of the object, and (b) the te the accelerationstring. of the object and (b) the tension in the string. Picture the Problem The force diagram shows the forces acting responsible acceleration ofSolution the sphere and the difference between the weight o the tension is the net force acting on the hanging object. We can us We can draw the force diagram for the syslaw to obtain two equations in a and T that we can solve simultane ———————————————————————————————————— and the hanging object. The tension in the string is (a) tem 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, X τ = T R = Iα. F T R 0 M While for the hanging mass, calling the vertical direction x, X Fx = T − mg = −ma. T mg x x ∑ Now, since the string isn’t slipping wheel, the angular acceleration (a) Noting that Taround = T ′, the apply τ =2TRis= 2 0 just α = a/R. Furthermore, the moment of inertia for a sphere is I = M R . So, nd 5 Newton’s 2 law to the sphere and we find the following system of equations and the hanging object: 2 T = Ma 5 T = m (g − a) . ∑F x I sphereα = mg − T = ma ( Setting these twoSubstitute expressionsfor equal to each acceleration Isphere andother α inand solving for the 2 2 = TR MR 5 gives equation (1) to obtain:g a= . 1 + 2M 5m ) Ra Eliminate T between (2) T = 2 M a to findg (b) Now we just substitute this result back inequations to the expression 5 and (3) and solve for a to obtain: a= 1+ 2 Mg 2M mg T = = . 2M 5 1 + 5m 5m + 2M (b) Substitute for 7 a in equation (2) and solve for T to obtain: T= 2M 5m 2mMg 5m + 2 M 8. A basketball rolls without slipping down an incline of angle θ. The coefficient of static friction is µs . Model the ball as a thin spherical shell. Find Picture Problem three (a) the acceleration of the the center of mass of The the ball, forces acting on the basketball a the ball, theonnormal force, and the force of friction. Because th (b) the frictional force acting the ball, and assumed to be acting at the the normal force (c) the maximum angle of the incline for which thecenter ball willof rollmass, withoutand slipping. 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 Solution law to find a system of simultaneous equations that we can solve f (a) We can begin by writing Newton’s statement. called for indown the problem laws for P P Fx PFy τi y the ball, including the torques: = −mg sin θ + Ff = −ma = −mg cos θ + Fn = 0 = Ff 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 Ff = Iα/R = Ia/R2 to the first equation and solving for a gives a= r Fn m x r 0 r mg θ mg sin θ . m + I/R2 nd r f ∑ ∑F (aof) Apply law in both The moment inertia ofNewton’s the spherical2shell is just I = 23 mR2 , and soFx translational and mg rotational form to sin θ 3 a= = g sin θ. m + 2m/3R2 5 the ball: (b) Since Ff = I a R2 = 2 mR2 3 R2 a = 23 ma, we have 2 Ff = mg sin θ. 5 y = mg sin θ − f s = = Fn − mg cosθ = and ∑τ 0 = f s r = I 0α a this Because basketball (c) Now, we know that thethe frictional force isisFfrolling = µs FN = µs mg cos θ. Setting α = equal to ourwithout result from part (b) we we find r slipping know that: 2 5 mg sin θ = µs mg cos θ ⇒ tan θ = µs . 5 2 a Substitute in equation (3) to So, the maximum angle of the incline that the ball will roll without f s rslipping = I 0 is r obtain: 5 −1 θ = tan µs . 2 Table 9-1 we have:from the sliding case. The rolling From of the ball increases the angle 8 α in equation Substitute for I0 and (4) and solve for fs: I 0 = 23 mr 2 fs r = ( 2 3 mr 2 ) ar ⇒ f s = 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 L0 from the end of the ramp. Find the ratio L0 /L. ———————————————————————————————————— Solution 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 L0 = v 0 ∆t, where v is the speed of the shell and v 0 is the speed of the sphere at the bottom of the ramp. Now, the ratio L0 /L = v 0 /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, 1 mv 2 + 12 Ishell ω 2 2 1 mv 02 + 12 Isphere ω 02 2 KEshell = KEsphere = Now, since the balls are rolling without slipping, ω = v/R, while ω 0 = v 0 /R. Plugging in these expressions gives 2 shell KEshell = 21 1 + ImR 2 mv Isphere KEsphere = 12 1 + mR mv 02 2 Since energy is conserved KEshell = KEsphere = mgH. So, 2 shell mgH = 12 1 + ImR 2 mv Isphere mgH = 12 1 + mR mv 02 2 Taking the ratio v 0 /v gives v u u 1 + Ishell2 v mR =t . Isphere v 1 + mR2 0 Since Ishell = 23 mR2 , while Isphere = 25 mR2 , we have s r 0 0 v L 1 + 2/3 25 = = = = 1.09. v L 1 + 2/5 21 9 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 10−18 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 1.00 × 10−18 m, what angular speed would be necessary to produce the observed intrinsic angular momentum of ~/2? (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? ———————————————————————————————————— Solution (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 = 25 M R2 , and for L = ~/2, we find ω= 5~ L = . I 4M R2 Taking M = me = 9.11 × 10−31 kg, we have ω= 5 × 1.05 × 10−34 5~ = = 1.44 × 1032 rad/sec. 2 −31 −18 2 4M R 4 × 9.11 × 10 × (10 ) (b) The speed is v = Rω, so using our results from part (a) gives v = Rω = 10−18 × 1.44 × 1032 = 1.44 × 1014 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