PY1052 Problem Set 7 â Autumn 2004 Solutions

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where I com is the moment of inertia for rotation about an axis that is parallel to theone in which we are interested and passes through the centre of mass, and h is thedistance between the COM axis and our axis. In our case, I com is just equal to I mid ,since the axis for I mid is parallel to the one we are considering and passes throughthe metre stick’s centre of mass. The distance h is just equal to the distance betweenthe middle of the metre stick (the 0.50 m mark) and our axis (the 0.20 m mark), or0.30 m. Therefore:I = I com + Mh 2 = 0.047 kg m/s + (0.56 kg)(0.30 m) 2 = 0.097 kg m/s(6) A delivery truck operates by using the energy stored in a rotatingflywheel, which is a solid cylindre with a mass of 500 kg and a radius of1.0 m. The truck is charged by using an electric motor to get the flywheelup to its top speed of 200π rad/s. (a) What is the kinetic energy of theflywheel after charging? (b) If the truck operates with an average powerrequirement of 8.0 kW, for how many minutes can it operate betweenchargings (1 W = 1 Watt = 1 J/s)?(a) The kinetic energy of rotation of the flywheel isKE = 1 2 Iω2We know that ω = 200π rad/s, and must determine the moment of inertia I of theflywheel to find its kinetic energy. Your table indicates that the moment of inertia ofa solid cylinder or disk about its central axis isI = 1 2 MR2 = 1 2 (500 kg)(1.0 m)2 = 250 kg m/sTherefore, the kinetic energy isKE = 1 2 (250 kg m/s)(200π rad/s)2 = 4.93 × 10 7 J(b) The average power is given byP avg = WorktimeHere, the work done between when the wheel is fully charged (rotating at its maximumangular speed) and uncharged (when it no longer rotates) is equal to the change inthe wheel’s kinetic energy. The time is what we are looking for. Thereforetime = KEP avg= 4.93×107 J8.0×10 3 W = 6.16 × 103 s = 103 min(7) A thin spherical shell has a radius of 1.90 m. An applied torqueof 960 N m gives the shell an angular acceleration of 6.20 rad/s 2 about anaxis passing through the centre of the shell. (a) What is the moment ofinertia of the shell? (b) What is the mass of the shell?(a) You can find from a table of moments of inertia that the moment of inertia of athin spherical shell about any diameter isI = 2 3 MR2but in this situation, we can’t use this directly, because we don’t know the mass ofthe shell. Instead, use Newton’s second law for rotation:τ = Iα −→ I = τ αI =960 N m = 155 kg m/s6.20 rad/s 2(b) Now, using the expression for I for a spherical shell and the answer to (a), wecan find the shell’s mass:I = 2 3 MR2M = 3I2R 2=3(155 kg m/s2(1.90 m) 2 = 64.4 kg
(1b) A 6100 kg rocket is set for vertical firing from the ground. If theexhaust speed is 1200 m/s, how much gas must be ejected each second ifthe thrust is to (a) equal the magnitude of the gravitation force on therocket and (b) give the rocket an initial upward acceleration of 21 m/s 2 ?(a) The thrust is given by Rv rel , where R is the rate at which the fuel is consumedand v rel is the velocity of the rocket relative to the ejected gases. In order for thethrust to be equal to the gravitational force on the rocket,Rv rel = Mg −→ R = Mgv relR = (6100 kg)(9.8 m/s2 )1200 m/s= 50 kg/s(b) In order for the rocket to have an upward acceleration of 21 m/s 2 ,Thrust − Mg = Ma −→ Thrust = M(g + a)Rv rel = M(g + a)R = M(g+a)v relR = (6100 kg)((9.8+21) m/s2 )1200 m/s= 156 kg/s(2b) A pulsar is a rapidly rotating neutron star that emits a radio beamin much the same way that a lighthouse emits a light beam. We receivea radio pulse once each time the star rotates, and we can measure theperiod T of the rotation by measuring the time between the pulses. Thepulsar in the Crab Nebula has a period of rotation of T = 0.033 s, and thisperiod is increasing at a rate of 1.2 × 10 −5 s/year. (a) What is the pulsar’sangular acceleration? (b) The pulsar was born in a supernova explosionseen in the year 1054. What was the initial period T of the pulsar if itsangular acceleration has been constant since the time it was born?The relationship between the period and angular velocity isT = 2π = ω2πω−1Taking the time derivative of both sides:dTdtdω= −2πω−2= dt −2πω−2 α−2 dT= −2πTα = − ω2 dT2π dtdtWe are given that T = 0.033 s and dT/dt = 1.2 × 10 −5 s/year:dTdtTherefore,s= 1.2 × 10−5 × 1 year × 1 day × 3600 s = 3.8 × 10 −13 s/syear 365.25 days 24 hrs 1 hrα = −2π(0.033 s) −2 (3.8 × 10 −13 s/s) = −2.2 × 10 −9 rad/s 2The negative sign indicates that the ω is decreasing, which makes sense because theperiod is increasing.(b) If the angular acceleration is constant, dT/dt is constant. There are 949 yearsbetween 1054 and 2003, soT 2003 = T 1054 + dTdt ∆t −→ T 1054 = T 2003 − dTdt ∆tT 1054 = 0.033 s − (1.2 × 10 −5 s/year)(949 years) = 0.033 s − 0.011 s = 0.022 s
Page 2 and 3: p cue,i = 4.75m kg m/s îp cue,f =

PY1052 Problem Set 7 â Autumn 2004 Solutions

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PY1052 Problem Set 7 â Autumn 2004 Solutions