The acceleration due to gravity g, which varies with latitude and height, will be approximated to 101m1s 2, unless otherwise specified. F1 (a) The fulcrum in this case is the horizontal axis, perpendicular to the page, passing through the elbow joint at B. (b) Figure 18 shows the forces on an idealized diagram. R is the reaction at the elbow joint, T is the tension in the biceps muscle, W is the weight of the arm and L is the weight of the object. Without the force R, it would be impossible to achieve rotational equilibrium. This is seen most easily by taking moments about the centre of mass of the arm. The forces T and L from the biceps muscle and the x z B R T Figure 183See Answer F1. a b A W = 1 101N c L = 2 101N object both promote clockwise rotation. This can only be balanced if there is a downwards force R at the elbow joint. Since none of the other forces have horizontal components the force R must act vertically downwards in order to ensure translational equilibrium. y
(c) For translational equilibrium: z T R W L = 0 For rotational equilibrium, taking moments about the elbow (point B0), and taking anticlockwise moments to be positive: at bw c0l = 0 (d) Suppose a = 41cm, b = 161cm, c = 321cm and the weight of the forearm W = 1 101N, where the magnitude of the acceleration due to gravity is rounded up to 101m1s 2. We have L = mg = 2 101N and the translational equilibrium condition becomes: x B a R T Figure 183See Answer F1. T R = 301N The rotational equilibrium condition about B does not involve R and becomes: T (0.041m) = 101N (0.161m) + 201N (0.321m) = 8.01N1m b A W = 1 101N c L = 2 101N y
i.e. T = 8.01N1m/(0.041m) = 2001N z Substituting for T in the first equation then gives R = 1701N. (e) Figure 18 shows a right-handed Cartesian coordinate system with origin at the elbow. The general expression for the G torque of a force F applied at position r G is = r1 1F. Here we need the torque from L about B. This is: = (0.321m)1j1 1(201N)( k) = 0.32 20(1j1 1k)1N1m G = 6.4i1N1m x B a R T Figure 183See Answer F1. b A W = 1 101N c L = 2 101N This torque vector points along the negative x-axis. This tells us that someone on the positive side of the x-axis, looking towards the origin, will see an associated clockwise sense of rotation associated with this, as expected. y
F2 (a) The moment of inertia of the column about the central axis is I = Mr 2 /02 = πr 2 lρ0r 2 /02 = πr0 4 lρ/02 = π(0.751m) 4 401m 2.4 10 3 1kg1m 3 /02 = 4.8 10 4 1kg1m 2. (b) Using the parallel-axis theorem I = Mr0 2 /02 + Mr 2 = 3Mr0 2 /02 = 1.4 10 5 1kg1m 2 (c) Figure 19 shows a side elevation of the rolling column. The G torque about the line of contact through A, due to the weight W, is = r1 1W Figure 193See Answer F2. G which has a direction out of the plane of Figure 19 (towards you) z 10 10 r d A W instantaneous axis of rotation x
and a magnitude G G w a Γ = rw1sin1170 = r πr 2 lρ0g 1 sin1170 = πr 3 lρ0g sin1170 = π(00.751m) 3 401m 2.4 10 3 1kg1m 3 101m1s 2 0.174 = 2.2 10 5 1N1m (d) From the basic law of rotational dynamics 10 for uni-axial motion = Ia. The magnitude of r the angular acceleration is α = Γ0/I = 2.2 10 5 1N1m/1.4 10 5 1kg1m 2 = 1.61rad1s 2. d A The direction of the angular acceleration vector is along (i.e. out of the page, towards you). (e) Since the angular acceleration is constant 10 W and the initial angular velocity is 0, the final angular velocity is = a0t. Thus the magnitude of the angular velocity vector is Figure 193See Answer F2. ω = 1.61rad1s 2 21s = 3.21rad1s 1 and its direction is along (out of the page, towards you). z instantaneous axis of rotation x
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to Ready to study? in Subsection 1.3. Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Closing items. If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.
R1 Since the arc of a circle = radius angle in radians, the distance moved by the mountain is rθ = (6.38 10 61 m) 2 π/0180 = 2.23 10 51 m. If the angular speed is ω and the period of the Earth s rotation is T = 241hr, then ω = 2π/Τ = 2π/(24 60 60)1s = 7.27 10 5 1rad1s 1 and the speed is v = rω = 6.38 10 61 m 7.27 10 5 1rad1s 1 = 4641m1s 1
R2 The component of a force along a given direction is given by the product of the magnitude of the force and the cosine of the angle between the force and the direction concerned. From Figure 2, the force is at an angle of 30 to the positive x-direction and 60 to the negative z-direction. The x-component of F F x = F1cos130 = (121N) cos130 = 10.41N The z-component of F is F z = (121N) cos160 = 6.01N Newton s second law of motion is F = ma, which leads to a x = F x 0/m = 10.41N/3.01kg = 3.51m1s 2 and a z = F z /m = 6.01N/3.01kg = 2.01m1s 2 z A 60 F = 121N O Figure 23See Question R2. x
R3 We solve this problem by subtracting a smaller from a larger imaginary solid pipe, having the given inner and outer radii, respectively. Since mass = density volume, it is necessary first to calculate the volumes of these solids. Do not forget to divide the diameter by 2 in order to calculate the radius. The volume of a cylindrical solid = cross-sectional area height = πr 2 h. The volume of the larger pipe is, therefore, π(0.401m) 2 241m. The volume of the smaller pipe is π(0.371m) 2 241m. The mass of the hollow pipe is, therefore, density volume = 2.4 10 3 1kg m 3 (0.40 2 0.37 2 )1m 2 241m π = 4.2 10 3 1kg The weight of the pipe acts downwards and has magnitude mg = 4.2 10 3 1kg 101m1s 2 = 4.2 10 4 1N
T1 Suppose your weight is 7001N and that this is just sufficient to loosen the nut when you stand on a spanner of length 0.41m, fitted horizontally. The size of the torque due to your weight is: Γ = 7001N 0.401m = 2801N1m. This must also be the maximum frictional torque on the nut.
T2 The friction of the ground acting on the tyre prevents the wheel from slipping. As you begin to jack up the wheel the maximum friction available reduces and the wheel is more likely to slip1 1this is why the wheel nuts should be loosened slightly before the wheel and tyre are lifted off the ground.
T3 Figure 20 shows the tension force, T, and one of the tie cords, PQ, in side elevation. Using Equation 1, torque magnitude33γ = r1f1sin1θ = Fd (Eqn 1) the magnitude of the torque about O is: Γ = Td = rt1sin1θ = 1201N 41m [81m/(4 2 + 8 2 ) 1/02 1m] = 4291N1m x z P 4 m O θ d T 8 m Figure 203See Answers T3, T4, T7 and T8. This shows a mast with one of its tie cords. Q y
T4 We can refer again to Figure 20. (a) If the fulcrum is taken to be on the mast 11m above the ground, then it is 31m below the point of attachment of the cord at P. The torque about this fulcrum is calculated as in Answer T3, but with 31m replacing the 41m in that calculation (θ is the same): Γ00 = Td0 = r Tsinθ = 1201N 31m [81m/(4 2 + 8 2 ) 1/02 1m] = 3221N1m (b) If the fulcrum is taken to be coincident with the point of attachment of the cord at P then the distance of the line of action of the force from the fulcrum is zero and so the torque is zero. x z P 4 m O θ d T 8 m Figure 203See Answers T3, T4, T7 and T8. This shows a mast with one of its tie cords. Q y
T5 Equation 2b W = Γ11 1φ1 (Eqn 2b) gives the work done as the average torque multiplied by the angle (in radians). The angle turned through is one revolution (2π radians) and the average torque Γ = 2801N1m/4 = 70.01N1m. So, W = 701N1m 2π = 4401J. If this is completed in 51s, the average power is given by Equation 3 P = W t = Γ φ t as P = W/ t = 4401J/51s = 88.01W. = Γω (Eqn 3)
T6 Figure 21 shows an idealized version of the balance. Suppose the point of support is at a distance p (in metres) from the line of action of the larger weight. The 201kg mass has a weight of 201kg 101m1s 2 = 2001N. Similarly the weight of the 31kg mass is 301N. The magnitude of the moment of the 2001N weight about the point of support is (200p)1N1m and since this moment promotes an anticlockwise rotation we take it to be positive. The magnitude of the moment of the 301N weight is 301N [1.5 (0.3 + p) 0.1]1m = 30 (1.1 p)1n1m 0.3 m 2001N Figure 213See Answer T6. Since this moment promotes a clockwise rotation we take it to be negative. Rotational equilibrium requires that the resultant moment be zero so (200p)1N1m [30 (1.1 p)]1n1m = 0. Hence p = 331N1m/0(2301N) = 0.141m. Hence the point of support is 0.141m from the line of action of the larger weight. p 1.5 m 301N 0.1 m
T7 z From Figure 20 we see that the tension T lies in the y-z plane, with no x-component. The y-component is T1sin1θ and the z-component is T1cos1θ. In terms of unit vectors we have: T = T1sin1θ1 j T1cos1θ 1k, where T is 1201N and θ is 63.43, so this gives: T = (107.31j 53.671k)1N. x P 4 m O θ d T 8 m Q y Figure 203See Answers T3, T4, T7 and T8. This shows a mast with one of its tie cords.
T8 (a) The coordinate axes are those shown in Figure 20; the position vector r of the point P of attachment of the cord to the mast is: r = 4k1m. (b) The torque is then: ( )Nm = r T = 4k 107.3 j G 53.67k = 429( k j)nm = 429i Nm This confirms that the torque is along the negative x-axis, as expected from the corkscrew rule or the right hand rule, with a size as given in Answer T3. (i.e. 4291N1m) x z P 4 m O θ d T 8 m Figure 203See Answers T3, T4, T7 and T8. This shows a mast with one of its tie cords. Q y
T9 In each revolution of a wheel the distance travelled by the car is 2πr, where r is the radius of the wheel. The speed of the car is 100 10001m/36001s = 27.81m1s 1. (a) This speed of 27.81m1s 1 must also be the speed of a point on the circumference. (b) The point on the wheel which is in momentary contact with the ground is being carried forward with the overall speed of the vehicle but has an opposite motion due to its rotation1 1so it is momentarily at rest relative to the road. You may find this surprising, but the tyres would soon wear out if it weren t true! (c) We use v = rω to find ω = v/r = 27.81m1s 1 /0.351m = 79.41rad1s 1. (d) Figure 12 can be used to represent the situation, with the circular path becoming the wheel. Since w is along the z-axis, we have w = 79.41k1rad1s 1. y r θ Figure 123A particle moving in a circle. The (x, y) plane forms the plane of rotation and the z-axis points towards you. Looking down onto the (x, y) plane from the positive z-axis, the angle θ increases for anticlockwise rotations. P x
T10 (a) Since the wheel rotates about the z-axis we write the equations of uniform angular acceleration with z subscripts. Using θ z = (µ0 z + ω0 z )t/02 we obtain θ z = (79.4 + 0)1rad1s 1 31s = 2381rad (b) The kinematic equation ω z = µ z + α0 z t applies because the acceleration is uniform. Clearly ω z = 0 and µ z = 79.41rad1s 1 from Answer T9. So α0 z = 79.41rad1s 1 /6 s = 13.21rad1s 2. The magnitude of the angular acceleration is therefore 13.21rad1s 2. (c) To find the angular speed after 31s we can use the equation: ω z = µ0 z + α0 z t = 79.41rad1s 1 13.21rad1s 2 31s = 39.81rad1s 1.
T11 Observation 3 is inconsistent with the claimed equation. It tells us that the angular acceleration of the rod depends on the direction in which the torque acts, relative to the long axis of the rod. This does not agree with the suggested equation, which gives an acceleration of magnitude Γ/m irrespective of the direction of the torque.
T12 (a) Using Equation 12 = Ia = I dw G (Eqn 12) dt and taking the magnitudes only we have Γ = Iα and so I = Γ0/α. In terms of base quantities, the dimensions of torque Γ are those of force distance = [M][L][T 2 ] [L] = [M][L 2 ][T 2 ]. The dimensions of angular acceleration α are [T 2 ] since angular measure is dimensionless. Finally, the dimensions of moment of inertia I are [M][L 2 ][T 2 ]/[T 2 ] = [M][L 2 ]. (b) Suppose that the roulette wheel is initially spinning so that its angular velocity vector points along the upward z-axis. The initial angular velocity is then µ z = 2πf = 2π/02 = π1rad1s 1 and the final angular velocity is ω z = 0. The angular acceleration can be found using ω z = µ0 z + α0 z t which gives α0 z = µ0 z /t = π1rad1s 1 /121s = 0.2621rad1s 2 Finally, the retarding torque is given by Γ z = Iα0 z = 251kg1m 2 0.2621rad1s 2 = 6.551N1m. The magnitude of the torque is therefore 6.551N1m.
T13 (a) Since v = rω the kinetic energy is E kin = 1 2 mv2 = 1 2 mr2 1ω 2. The rate of change of kinetic energy is found by differentiating this expression with respect to time. Since m and r are constant and dω/dt = α, this gives de kin /dt = mr 2 ω α (b) The rate of change of kinetic energy must equal the power input so Equation 3 P = W t = Γ φ t = Γω (Eqn 3) gives de kin /dt = mr 2 ωα = Γω. Thus, Γ = mr0 2 α and comparing this with Equation 12 dt allows us to identify the moment of inertia as I = mr 2. G = Ia = I dw (Eqn 12)
T14 (a) Suppose the wheel has mass M and radius a and that I cm is its moment of inertia about the central axle. Using the parallel-axis theorem we then find I contact = I cm + Ma2 = Ma 2 /02 + Ma2 = 3Ma 2 /02 which gives I contact = 3Ma 2 /02 = 3 121kg (0.31m) 2 /02 = 1.621kg1m 2 (b) The radius of gyration k is defined by: I = Mk 2 = 1.621kg1m 2 and so k = (I/M) 1/02 = (1.621kg1m 2 /121kg) 1/02 = 0.3671m. The moment of inertia of the wheel about the line of contact of the wheel on the ground is the same as if the mass of the wheel were concentrated 0.3671m from this axis. (c) The total kinetic energy (i.e. rotational motion about the contact axis or the sum of translational centre of mass motion and rotational motion about the centre of mass) is given by: E kin = I contact ω 2 /02 = Mk 2 ω 2 /02. The angular speed is ω = 2πf = 18.851rad1s 1 so the kinetic energy is: E kin = Mk 2 ω0 2 /02 = 121kg (0.3671m) 2 (18.851rad1s 1 ) 2 /02 = 2871J