Gravitational potential energ m1 Consider a rigid bod of arbitrar shape. We want to obtain a value for its gravitational potential energ. O r1 1 x The gravitational potential energ of an assembl of N point-like objects of masses m1, m2,... mn, is given b the sum of the gravitational potential energies of each and ever one of them. Imagine therefore to dissect the rigid bod into its man elementar constituting point-like objects. Take for definiteness the first one. The potential energ associated to this first point is m1g1 with respect to a given point taken as the zero of gravitational potential energ (i.e., ground level). On adding contributions from all points, we have: U = m1 g1 + m2 g2 + m3 g3 +... + mn gn = Mg cm The gravitational potential energ of a rigid bod can be obtained b replacing the rigid bod with a single point, with the mass of the whole bod, positioned at the center of mass.
Application: the pendulum L/2 x L/2 + x This is a phsical setup of considerable importance, relevant to a number of practical situations, including sports. A thin, uniform rod of length L and mass m, is pivoted about one end. The rod is released from rest in a horizontal position, and allowed to swing downward. Air resistance is neglected. Initiall, the energ of the rod is entirel potential (i.e., gravitational), while when the rod is perfectl vertical it is entirel kinetic (rotational) Question: what is the difference in gravitational potential energ between the initial configuration in which the rod is perfectl horizontal, and the final one, in which it is perfectl vertical? Imagine dividing the rod in man small pieces of mass δm. Each segment can be paired up with an identical one, right across from the centre of the rod. The net difference in potential energ for these two pieces, between horizontal and vertical configurations, is δmg [(L/2 + x) + (L/2 x)] = δmg L. On adding together contributions from all small segments, and dividing b 2 in the end (we do not want to double count), we get Ui Uf = mgl/2 (center of mass)
Example: A solid sphere with radius R = 8.5 cm, is released from rest and rolls down a ramp, dropping through a vertical height h = 0.61 m. The ball leaves the ramp moving horizontall. Through what horizontal distance does the ball move before landing, if the ramp is at a height L = 1.22 m above the floor? Energ conservation: Mgh 1 2 I = Mvcm (1 + 2 MR 2) h vcm but for a solid sphere I = 2MR 2 /5 ielding vcm = 10 gh L 7 at the bottom of the ramp The time that the sphere spends in the air after leaving the ramp is given b gt 2 /2 = L i.e., t = (2L/g) In that time, the sphere travels horizontall vcm t = 20 Lh = 1.46 m 7 Question: How man revolutions does the ball make during its fall? The angular speed of the ball when it leaves the ramp is ω = vcm / R In the time t that it takes for it to fall, it spins ωt radians, i.e., (vcm t /R) = 1.46 = 17.1 rads 0.085 Since each revolution is worth 2π rads, the total number of revolutions is 17.1/(2π) = 2.7
Rotational Dnamics and Static Equilibrium
Vector (cross) product Consider two vectors, A and B. The vector product of these two vectors is defined as the vector C = A B B θ A Important: Unlike the dot product, the cross product is a vector. It has a magnitude and a direction. Magnitude of C : C = A B = A B sin θ Direction of C : perpendicular to the plane containing both A and B Imagine sweeping A toward B. note difference with respect to dot product -- sine here A B A counterclockwise C comes out of the screen B clockwise C goes into the screen
Torque Consider a force F applied at a specific point P in space. Let r be the vector connecting the chosen origin of a sstem of coordinates to the point in which the force is applied. F out of the screen O θ r x P Definition: The torque associated to the force F with respect to the center O, is given b τ = r F τ = r F sin θ Question: which direction? In or out of the screen? Torque is a vector quantit. Its dimensions are [F][L] same as energ, but no specific unit is introduced for torque and one does not use J, in order not to confuse with energ. Example: Two helmsmen disagree on which wa the ship should turn. The exert the forces F1 and F2 shown. The magnitude of F1 is 72 N, that of F2 is 58 N. Find the net torque on the helm. The wheel has a radius r = 0.74 m. Torque of force F1: r1 F1 sin 50 o = (0.74)(72)(0.77) Nm = 40 Nm 90 o 50 o out of the screen Torque of force F2: into the screen Net Torque: 3 Nm into the screen r2 F2 sin 90 o = (0.74)(58)(1.0) Nm = 43 Nm weaker force wins out on more favourable angle r2
What does the torque do? The main effect of a torque is that of imparting a rigid bod an angular acceleration. To see how that comes about, consider the following simple example: a r 90 o Top View x Sphere of mass m connected to an axis of rotation (coming out of the screen) b means of a rod of negligible mass and length r The whole sstem can rotate in-plane Let a constant tangential force F be applied on the sphere Torque associated to that force: τ = r F sin θ = r F It is, of course, F = ma r F = mar But the acceleration a is tangential, as the sphere keeps moving on a circular path, held on it b the rod a = α r Hence, r F = m α r 2 or, τ = (m r 2 ) α τ = I α as mr 2 is the moment of inertia I of the sstem formed b the rod and the sphere. Although it was shown for a particular case, the relation obtained is valid in general In the presence of more torques, one has Σi τi = I α for a rigid bod
Example: A light rope is wrapped around a disk-shaped pulle is pulled tangentiall with a force of 0.53 N. Find the angular acceleration of the pulle, given that its mass m is 1.3 kg and its radius 0.11 m. r τ = r F sin θ = r F sin θ = 1 as θ = π/2 (aka 90 o ) m F Moment of Inertia of disk is I = mr 2 /2 2F 2 0.53 τ = I α hence, r F = (mr 2 /2) α α = = = 7.4 rad/s2 mr (1.3)(0.11)s2 Example: The L shaped object shown here consists of three masses connected b light rods. What torque must be applied to this object to set it on rotation at an angular speed 1.2 rad/s 2, if it rotates about the x-axis? 9 kg I = (9 kg)(1 m 2 ) τ = I α = (1.2)(9) Nm = 10.8 Nm 1m 2m 1.2 kg 2.5 kg x What if the rotation takes place about the -axis? I = (2.5 kg)(4 m 2 ) = 10 kg m 2 What if the rotation takes place about the z-axis (out of the screen)? τ = I α = (10)(1.2) Nm = 12 Nm I = (2.5 kg)(4 m 2 ) + (9 kg)(1 m 2 ) = 19 kg m 2 τ = I α = (19)(1.2) Nm = 22.8 Nm
Pulle Let us now tr and understand in detail how an actual pulle works. A pulle is essentiall a disk-shaped wheel, with a mass m and a radius r. Its moment of inertia computed with respect to an axis going through its centre is therefore I = mr 2 /2 r T The pulle, as we have seen, is usuall utilized to lift, or gentl lower weights, b having a string wrap around it. Consider a such a situation, with a crate of mass M being lowered. T The string ought not slip over the disk, i.e., there has to be friction between the string and the disk W M Forces on the crate: Weight W thus, T Mg = Ma Tension T (crate descends) Pulle is tpicall firml mounted on the ceiling and does not move. There is, however, a torque on it exerted b the string. Torque on the pulle: it is eas to see that τ = r T but, τ = I α hence, r T = mr 2 α /2 String move together with both pulle and crate, i.e., a = α r i.e., T = ma /2 Therefore, from crate force equation we get, on substituting T, g a = 1 + (m/2m)
Static Equilibrium Definition: A sstem that is capable of both translational as well as rotational motion is in static equilibrium if both the following conditions are met: Σi Fi = 0 (net external force) Σi τi = 0 (net external torque) If onl one of these two conditions is met, sstem is not in static equilibrium These two conditions are generall independent of one another Example: A child of mass m is supported on a light plank b his parents, who exert forces F1 and F2. Find the values of these forces so that the plank is in static equilibrium. Net Force : (F1 + F2 mg) = 0 ( direction onl) Net Torque : Torque of force F1 = 0 Torque of force W = Torque of force F2 = L F2 3mgL/4 = 0 (distance from origin is zero) (3L/4) mg L F2 Torque condition ields F2 = 3mg/4 (into the plane) (out of the plane) Force condition ields F1 + 3mg/4 mg = 0 F1 = mg/4 (tr for ourselves placing the origin where, e.g., the child is. The answer will not change)
Example: An 85-kg person stands on a lightweight ladder. The floor is rough, while the wall is frictionless. Find the magnitudes of all forces exerted on the ladder. B Q O Fs L = 3.8 m W x d = 0.7 m Forces on the ladder: Normal force from floor Q Friction force from floor Fs Weight of person W Normal force from wall B Net Force : Q + F + W + B = 0 s along x: Fs B = 0 Fs = B along : Q mg = 0 Q = mg = 834 N Net Torque : Torque of force Q = 0 Torque of force Fs =0 Torque of force W = Torque of force B = BL mgd = 0 B = mgd/l = (0.7 Q)/3.8 = 154 N (distance from origin is zero) (distance from origin is zero) mg r sin (90 o + θ) = mg r cos θ = mgd (into the plane) B R sin (180 o θ) = B R sin θ = BL (out of the plane)