As shown in the sketch, a 17 kg, 3.80 m pole is balanced vertically on its tip.
ID: 2203196 • Letter: A
Question
As shown in the sketch, a 17 kg, 3.80 m pole is balanced vertically on its tip. Then it starts to fall with no slipping at its lower end. What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.] m/s If on the upper end of the pole there is a ball which has a mass of Mpole/3, what is the speed of the ball (or the upper end of the pole) just before it hits the ground? (Assume the radius of the ball can be ignore compared with the pole) m/sExplanation / Answer
Nearly right, Ivan A, but the formula I = (1/3)mL² is for a pole of length 2L, not L (see Introduction to the Theory of Mechanics, 1965, by K.E. Bullen, p.180, agreeing with my own calculation, maybe because he taught me in 1957!!) So we can use that formula with L = 1.90 In that case the initial gravitational potential energy is mgL Strictly speaking, the question should say that the pole is of uniform linear density, but we assume that's the case otherwise the problem hasn't enough info to be solved. Another thing to take into account is that there's linear Kinetic Energy too, because the centre of mass of the pole is moving with speed Lw (still taking the length of the pole as 2L), and so total K.E. = (1/2)( Iw² + m(Lw)²) ............. = (1/2)*(4/3)mL²w² ............. = (2/3)mL²w² I verified this, too, by considering the K.E. of an element of length ds at a distance of s from the end around which it's rotating. The K.E. is (1/2)?s²w² ds where ? is the density.and of course the linear speed is sw. Integrating from 0 to 2L gives the above result. Now solve (2/3)mL²w² = mgL to get w = v(3g/2L) The speed of the upper end is 2Lw = v(4L² * 3g/2L) ...... = v(6gL) ...... = v(6*9.8*1.9) ...... = 10.57m/s, or rather, since data is given correct to 3 figures, 10.6 m/s. 2. Does FT mean "force of tension"? Not being able to do subscripts here, I prefer to use F, and will do so. The horizontal acceleration towards the centre of the circle is a = rw², and in this case r = 0.6 cos? and w = 2.4p radian/s The horizontal component of the tension must be equal to ma (what does R mean? Oh yes, radial.), and so F cos? = 0.14*0.6 cos? *(2.4p)² therefore F = 0.14*0.6*(2.4p)² ............... = 4.7753 N [4.78 to 3 figures] The vertical component of the tension, F sin?, must be equal to the weight, since vertical acceleration = 0, and therefore F sin? = 0.14g Hence sin? = 0.14*9.8/F Therefore ? = 16.7 degrees correct to 3 figures. 3. Speed 88 km/h ............ = 88000/3600 m/s .......... v = 220/9 m/s Call the force of friction (in the direction of the track surface, i.e. it acts at 12° down from the horizontal) F. The normal reaction of the track surface is R. Then since vertical acceleration = 0 and horizontal acceleration = mv²/r we have R cos 12° - F sin 12° = 0 Therefore R = F tan 12° also R sin 12° + F cos 12° = 1900*(220/9)²/70 Substitute R = F tan 12° into this equation: F (sin 12° tan 12°+ cos 12°) = 1900*(220/9)²/70 And, since I'm too lazy to use a calculator, paste into excel the expression =1900*(220/9)^2/(70*(sin(radians(12))*ta… + cos(radians(12)))) to get F = 15864.27749, or 16000 N since all the data are given to 2 figures. {though we can't be sure about the mass since it isn't written as 19*10²)
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