BBrilliant STUDY CENTRE PHYSICS (ONLINE) f ≤ µsmg MF ≤ µsmg M+m MF ≤ (M + m)µsmg F ≤ m (M + m) µsg M If F > µs m (M + m) g, the relative sliding between the blocks starts. The friction force between the blocks M will be kinetic nature and both blocks will move with different acceleration. (m) ⇒ F − µKmg = ma1 a1 = F − µKmg m fK = Ma2 µKmg = Ma2 a2 = µKm g M 51
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Example : Two blocks of masses M and m are arranged as shown in fig. There is no friction between ground and block M. The coefficients of static and kinetic friction between M and m are µs andµk respectively. a) Calculate the maximum possible value of F so that both the bodies move together b) Find the accelerations of the blocks if F is greater than that found in part (a). Solution If both blocks move together ie no sliding between M and m, the friction between m and M will be static nature. Static friction force is a self adjusting force O < f ≤ µsN Acceleration of system in this case a = F − −(1) M+m Equation of motion of m f = ma = mF M+m If there is no sliding between M and m then friction is static f ≤ fms f = µsN = µsmg 52
BBrilliant STUDY CENTRE PHYSICS (ONLINE) mF ≤ µsmg M+m mF ≤ (m + M)µsmg F ≤ (m + M)µsmg m b) If F > µs (M + m)g then there will be relative sliding between M and m. When relative sliding between M and m starts then friction is kinetic For (m) ⇒ fK = ma1 µKN = ma1 µKmg = ma1 a1 = µKg For (M) F − µKN = Ma2 F − µKmg = Ma2 a2 = F − µKmg M 53
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Minimum mass hung from the string to just start the motion When mass m1 is placed on a rough horizontal plane another mass m2 hung from the string connected by frictionless pulley. The tension (T) produced will try to start the motion of mass m1 Motion T ≥ fms Limiting condition T = fms N = m1g fms = µs N = µsm1g (m2 ) ⇒ T = m2g − −(1) (m1) ⇒ T = µsm1g − −(2) m2g = µ2m1g m2 = µsm1 µs = m2 m1 54
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Maximum Length of Hung Chain A uniform chain of length is placed on the table in such a manner that ' part is hanging over the edge of table without sliding. Since the chain have uniform linear density therefore the ratio of mass and ratio of length for any part of chain will be equal. We know µ = m2 = mass hanging from the table m1 mass lying on the table m2 = m ' ( )m1 m = − ' µ = ' = length hanging from the table [As chain have uniform linear density] − ' length lying on the table By solving ' = µ (µ +1) Sticking of a block with accelerated cart 55
BBrilliant STUDY CENTRE PHYSICS (ONLINE) FBD of block m [Cart as a frame] N − ma = m × 0 N = ma Now the block will remain static with respet to cart if friction f ≥ mg µN ≥ mg µma ≥ mg a≥ g µ ∴ amin = g µ Example In amusement parks there is a device called rotor where people stand on platform inside a large cylinder that rotates about a vertical axis when the rotor reaches a certain angular velocity, the platform drops away. Find the minimum coefficient of friction for the people not to slide down. In this case normal reaction of surface provides centripetal force and friction force prevents the man 56
BBrilliant STUDY CENTRE PHYSICS (ONLINE) from sliding vertically N = mv2 = mrω2 ................. (1) r To prevent downward slipping f ≥ mg µsN ≥ mg µsmrω2 ≥ mg ω2 ≥ g µsr ωmin = g µsr Motion in a vertical circle Consider a body of mass m tied to one end of the string and made to rotate in a vertical circle of radius r as shown in figure. Let vA be the velocity of the body at its lowest position. Its velocity for any angular position θ is vp. 57
BBrilliant STUDY CENTRE PHYSICS (ONLINE) cos θ = OE L OE = L cos θ AE = h = OA − OE = L − L cos θ Using conservation of mechanical energy KEA + PEA = KE P + PEP 1 mv 2 + O = 1 mvP2 + mgh 2 A 2 v2A = vp2 + 2gh vp2 = v 2 − 2gL[1− cosθ] A vP = cos θ Tension in the string By Newton’s second law along radial direction at point p Tp − mgcos θ = m v 2 p L Tp = mgcos θ + mvP2 L Case I At lowest position θ = 0 cos 0 = 1 TA = mv 2 + mg(max imum) A L Case 2 At B θ = 90 cos90 = 0 TB = mv 2 B L vB2 = v 2 − 2gL A At highest position θ = 180 cos180 = −1 TC = m v 2 − mg[minimum] C L 58
BBrilliant STUDY CENTRE PHYSICS (ONLINE) ( )v 2 = v 2 − 2gL 1−− 1 = v 2 − 4gL C A A v 2 = v 2 − 4gL c A Minimum velocity required at lowest position to complete the circle TH ≥ O mv 2 − mg ≥ O C L v 2 − 4gL ≥ gL A v 2 ≥ 5gL A ( )vA = 5gL min If vA ≥ 5gL , particle moves in vertical circle 0 < vA ≤ 2gL , particle oscillates like simple pendulum in arc of semicircle 2gL < vA < 5gL particle will give circular path after completing quarter circle and before completing semicircle. Centrifugal force FBD of the block in inertial frame FBD of the block in noninertial frame 59
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Consider a block of mass m placed on a table at a distance r from its centre. Suppose the table rotates with constant angular velocity ω . Let us first analyse the motion of block relative to an observer on ground (inertial frame). In this frame the block is moving in a circle of radius r. It therefore has an acceleration v2 towards the centre. In this frame forces on block r weight Normal reaction frictional force by the table We have N = mg .................(1) Newton’s second law of motion f = mv2 − − − (2) r Now observer the same block in a frame attached with the rotating table. The observer here finds that the block is at rest. Net force on the block in this frame must be zero. The weight and normal balance each other but frictional force acts on block towards the centre of block. To make the resultant zero, a pseudo force must be assumed to act on the block radially outward and has a magnitude mv2 or mrω2 . r This pseudo force is called centrifugal force. F =vertical 0 N–mg=0 N=mg Fradius=0 f − mv2 = 0 ; f = mv2 r r Analysis of conical pendulum Consider a conical pendulum of length l. It is made to rotate about a vertical axis. Suppose string of the pendulum makes an angle θ with the axis In inertial frame of reference T sin θ = mrω2 − −(1) from fig. sinθ = r l 60
BBrilliant STUDY CENTRE PHYSICS (ONLINE) T cos θ = mg − −(2) Eqn. 1÷ 2 tan θ = rω2 = ω l sin θ gg ω2 = g tan θ = g sin θ = g sin θ cos θ sin θ cos θ ω= g l cos θ Circular turning on roads When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which will produce the required centripetal acceleration. If the vechicles travel in a horizontal circular path, this resultant force is also horizontal. The necessary centripetal force is being provided to the vehicles by following three ways 1) By friction only 2) By banking of roads only 3) By friction and banking of roads both By friction only Suppose a car of mass m is moving at a speed v in a horizontal circular arc of radius r. In this case, the necessary centripetal force to the car will be provided by force of friction f acting towards centre Thus, f = mv2 r Further, limiting value of f is µN or f = µN = µmg (N = mg) Therefore, for a safe turn without skidding mv2 ≤ f r or mv2 ≤ µmg r v ≤ µrg By banking of roads only Friction is not always reliable at circular turns if high speed and sharp turns are involved to avoid dependence of friction, the roads are banked at the turn so that the outer part of the road is some what lifted compared to the inner part. 61
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Applying Newton’s second law N sin θ = mv2 ..............(1) r N cos θ = mg ..............(2) Eqn. (1) / (2) tan θ = v2 rg v = rg tan θ By both friction and banking of roads Now let tendency of car is to slip UP (this corresponds to maximum speed) then friction will act down the elevation 62
BBrilliant STUDY CENTRE PHYSICS (ONLINE) Along vertical direction Ncos θ = mg + µsNsin θ Ncos θ − µsNsin θ = mg − −(1) Along horizontal direction Nsin θ + µsNcos θ = mv2 − − − (2) r Dividing equation (1) by (2) we have sin θ + µs cos θ = v2 cos θ − µs sin θ rg Again dividing numerator and denominator each by cosθ we get v2 max µs + tan θ ( )= 1− µs tan θ rg v2 = rg(tan θ + µs ) max 1− µs tan θ vmax = s 1− µs tan θ Motion of cyclist on circular road When a cyclist goes round a circular road a centripetal force is required. The frictional force between the tyre and ground provides the necessary centripetal force. To get rotational equilibrium he leans from vertical. Let θ be the angle, the contact force on tyres makes with vertical. Then to prevent skidding mac ≤ fLim 63
BBrilliant STUDY CENTRE PHYSICS (ONLINE) mv2 ≤ µsN r mv2 ≤ µsmg r v ≤ µsrg Again we have fLim = mv2 − −(1) r N=mg---(2) Dividing equation (1) by (2) we get flim = v2 N rg fLim = tan θ N tan θ = v2 rg 64
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