PHYSICS LOCUS / / d dt. ( vi) mass, m moment of inertia, I. ( ix) linear momentum, p Angular momentum, l p mv l I

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Benedict Tyler
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1 6 n terms of moment of inertia, equation (7.8) can be written as The vector form of the above equation is...(7.9 a)...(7.9 b) The anguar acceeration produced is aong the direction of appied externa torque. The magnitude of the produced anguar acceeration is directy proportiona to the magnitude of the externa torque and inversey proportiona to the moment of inertia of the body. The above reation ooks the transationa equation F ma. Here, you shoud not forget that is not an independent rue. t is derived from F = ma ony. We can estabish an anaogue between transationa and rotationa variabes. By doing so concept deveoped so far for transationa motion woud hep to sove the probems invoving rotationa motion. The possibe anaogue is as foows / / ( i) distance traversed, S ange turned, ( ii) average speed, v s average anguar speed, t t ( iii) intantaneous speed, v ds d instantaneous anguar speed, ( iv) average acceeration, a dv average anguar acceeration, ( v) instantaneous acceeration, a dv instantaneous anguar acceeration, ( vi) mass, m moment of inertia, ( v ii) force, F torque, ( viii) F ma ( ix) inear momentum, p Anguar momentum, ( x) p mv dp d ( xi) F ( xii) conservation of inear momentum: conservation of anguar momentum: ( xiii) When F, p constant. When, constant. transationa kiic energy, k mv k rotationa kiic energy, ( xiv) work done, d F ds work done, d d t d Anguar quantities invoved in anaogues (ix) to (xiv) woud be discussed ater in this topic. Find the anguar acceeration of the rod given in exampe 4 at the moment (a) when it is reeased from rest in the horizonta position; (b) when it makes an ange with the horizonta.

2 7 After the moment when the rod is reeased from the rest in the horizonta position, it woud rotate in the vertica pane about a horizonta axis passing through the hinge and perpendicuar to the ength of the rod. nitia anguar veocity of the rod is zero but due to nonzero torque of gravity it has some anguar acceeration and hence, it wi acquire some anguar speed as it rotates. As discussed in exampe 4, the hinge force does not provide any torque about the axis under consideration and the weight of the rod tries to rotate it in the cockwise sense, i.e., it provides a torque perpendicuary, inward to the pane of the paper. An approach using r F to find the torque woud aso give the same resut. Hence, anguar acceeration of the rod, (b) m gravity g When the rod makes an ange with the horizonta, its anguar acceeration, r / cos m g cos [from figure 7.4 (b)] n the previous case, find the anguar veocity of the rod when it has turned through an ange after the moment when it was reeased from rest in the horizonta position. Aso find the anguar veocity when the rod becomes vertica. the rod, From the resut obtained in part (b) of the previous exampe, at some ange, the anguar acceeration of d g cos g cos d d g cos d [Using chain rue.] g. d cos d at =, =

3 8 g g sin sin When the rod becomes vertica, g sin g, and hence, anguar veocity, n the previous exampe, find the hinge force on the rod at =. Just after the moment when the rod was reeased from the rest in the horizonta position, it is shown in figure 7.4(a). Let the vertica component of the force on the rod from the hinge be R and the horizonta component of the same be R, as shown in figure. The subsequent motion of the centre of mass of the rod is a nonuniform circuar notice on the vertica circuar path of radius / with the centre at the hinge, as suggested in the figure. nitiay the rod is at rest and hence radia component of the acceeration of the centre of mass of the rod, ²r, is zero. Hence, appying F ma cm aong the radia direction, we get F ma, radia R m cm, radia Appying the same aong the tangentia direction, we have, =. Net hinge force F, tangentia R m ma R m 4 cm, tangentia 4 R R. 4 [ R ]

4 9 n the previous exampe, find the magnitude of the hinge force on the rod when the rod has turned through an ange. f be the anguar veocity of the rod when it has turned through an ange, the centre of mass of the rod has and as radia and tangentia components of its acceeration, respectivey, as shown in figure 7.4(b). Appying Fext Ma cm on the rod aong the radia direction, we have, R sin m R sin sin 5 sin Appying the same aong the tangentia direction, we have, cos R m R cos m cos cos 4 cos 4 Therefore, force on the rod from the hinge can be obtained by soving R R R. g sin = g = cos Rod when it makes an ange with the horizonta. R, R are perpendicuar and radia components, respectivey, of the reaction force acting on the rod from the hinge. You can aso assume reaction force as R and acting at some ange with the rod. f the disc given in exampe 5 has mass M and it is free to rotate about its symmetrica axis passing through O, find its anguar acceeration. f be the anguar acceeration of the disc, then, using, we have, FR ( MR ) 6F MR As the torque is in cockwise sense, has the same sense of rotation.

5 PHYSCS LOCUS 4 A uniform disc of radius. m and mass 5 kg is pivoted so that it rotates freey about its axis. A thin, massess and inextensibe string wrapped around the disc is pued with a force of N, as shown in figure 7.4(a). (a) What is the torque exerted on the disc about its axis? (b) What is the anguar acceeration of the disc? (c) f the disc starts from rest, what is the anguar veocity after s? t is obvious that the string force gives a torque to the disc in the cockwise direction. As the torque given by the force from the axe is zero. Net torque on the disc is, torque of the string force F r F R ( N) (. m).4 N-m. As the torque on the disc is in cockwise direction, the disc has anguar acceeration in the same direction. f be the magnitude of the anguar acceeration, ( MR ).4 rad/s 5 (.) rad/s At t = if the disc has zero anguar veocity, then, at some time t, its anguar veocity, in t t At t = s, rad/s A uniform disc of radius R and mass M is mounted on an axis supported in fixed frictioness bearing. A ight string is wrapped around the rim of the disc and a body of mass m is supported by the string, as shown in figure 7.4(a). (a) find the anguar acceeration of the disc; (b) find the magnitude of the tangentia acceeration of the point on the rim where the string separates from the rim. (c) if the system is reeased from rest at t =, find the speed of the bock at some time t (>).

6 4 Anayze the situation according to the information provided in the figure 7.4(b). You shoud aso note the foowing points: Ony tension force of the string, T, produces a torque on the disc about its centre O. Torque of the weight of the disc and that of the reaction force from the bearing are zero about O. f be the anguar acceeration of the disc (in the cockwise direction) then the point P on the disc has a tangentia acceeration R in the verticay downward direction at the moment shown in figure. The string unwinds at the same acceeration and the bock has the same acceeration in the verticay downward direction. Therefore, if a be the acceeration of the bock, then, a R...(i) Now, appying on the disc about its symmetrica axis, we have, Using F T. R T M MR R Ma T [Using (i)...(ii) ma for the bock in the vertica direction, we have, F ma T ma...(iii) Adding (ii) and (iii), we get, a m M m M m a g R f v be the speed of the bock at some time t, then, we have, v u at a is constant at u m m M gt Find the acceeration of m and m in an Atwood s Machine, shown in figure 7.44(a), if there is friction present between the surface of puey and the thread does not sip over the surface of the puey. Moment of inertia of the puey about its symmetrica axis is and its radius is R. The puey can rotate freey about its symmetrica axis.

7 PHYSCS LOCUS 4 Due to friction between the puey and the thread tensions in the parts of the thread on the two sides of the puey are different. Let that in the right part it is T and that in the eft part is T, as shown in figure 7.44(b). Forces acting on the two bocks and the puey are aso shown in figure 7.44(b). Force on the puey from the support and its weight are not shown because they do not produce torque on the puey about its symmetrica axis of rotation. f the bock m comes down with an acceeration a then m woud go up with the same acceeration because they are connected by the same string, as shown in the same figure.7.44(b). f we assume that the puey gets an anguar acceeration in the cockwise sense then the torque of T woud be positive and that of T woud be negative, as suggested in figure 7.44(c). Again, as any point on the rim of the puey has a tangentia acceeration R, the bock m comes down and the bock m goes up with the same acceeration, as shown in figure 7.44(d). Therefore, we can write, a Using Using F...(i) R ma for the two bocks, we have, mg T ma...(ii) [for m ] T m g ma...(iii) [for m ] for the puey, we have, T R T R...(iv) Substituting from (i) in (iv), we get, T T a R...(v) Adding (ii), (iii) and (v), we get, ( m m ) g a m m m m g m m R a R torque of support force and weight are zero

8 4 A thin uniform rod AB of mass m =. kg moves transationay with acceeration a =. m/s² due to two antiparae force F and F acting on it perpendicuary to its ength, as shown in figure The distance between the points at which these forces are appied is x = cm. Besides, it is known that F 5. N. Find the ength of the rod. Before anayzing the detais of the given situation, et us anayze the rotationa effect of two antiparae forces. Consider the situations shown in figure F and F are producing torques F and F are producing torques F and F are producing torques about A in opposite directions. about B in opposite directions. about C in the same direction f we anayze the torques of the two forces about every point in their pane containing them, then, we arrive at the concusion that if the point ies between the ines of action of F and F then torques of the forces about that point add up together otherwise they are in opposite directions. f the magnitudes of the two forces are equa then such a pair is caed as a coupe. f the magnitude of each force is F and the distance between their ines of appication is d, then, the torque about any point in their pane is F.d, as shown in figure F( d ) F F d F( d ) F F d Torque of a coupe. F d F d F( d d ) F d

9 44 Now, et us discuss the given case. As the rod is in pure transation motion, torque on it about any point must be zero. Therefore, the centre of mass of rod can not ie between the ines of action of the forces because in that case torques produced by then about the centre of mass do not cance each other. Let us assume that the centre of mass of the rod ies at a distance y away from the ine of action of F, as shown in figure As the rod transates towards right, F must have a greater magnitude than F. Using F ma, we have F F ma F F ma (5 ) N = N Again, as the torque on the rod about C must be zero, magnitude of the torque magnitude of the torque produced by F about C produced by F about C the two torques have opposite directions F ( x y) F y ( F F ) y F x y F x F F cm 5 = cm Length of the rod, ( x y). m A force F Aiˆ Bj ˆ is appied to a point whose radius vector reative to the origin of coordinates O is equa to r aiˆ bj ˆ, where a, b, A, B are constants, and iˆ, ˆj are the unit vectors of the x and y axes. Find the torque and the arm ength of the force F reative to the point O. Torque of F about O is r F ( aiˆ bj ˆ) ( Aiˆ Bj ˆ) ( ab ba) kˆ Arm ength of F with respect to O is r sin r is the distance of the point of appication of F from O and is the ange between r and F. r r F

10 PHYSCS LOCUS 45 ab ba a b a b A B ab ba A B A uniform cyinder of radius R is spun about its axis to the anguar veocity and then paced into a corner, as shown in figure 6.5(a). The coefficient of kiic friction between the corner was and the cyinder is equa to k. How many turns wi the cyinder accompish before it stops? A forces acting on the cyinder are shown in figure 6.5(b). As the cyinder rotates, its surface sips over the corner was and hence frictiona forces acting on it, f and f, are kiic in nature. Norma contact forces acting on the cyinder from the corner was, N and N, and the weight of the cyinder,, pass through the centre of the cyinder and hence, these forces produce no torque about the centre C. Ony frictiona forces produce torque about C and the torques produced by them are in opposite direction of the direction of the anguar veocity of the cyinder and hence, they retard the rotationa motion of the cyinder As the cyinder does not transate, force on it in both vertica and horizonta directions must be zero. Therefore, N and f N N N f...(i)...(ii) N N Substituting for N in equation (i) from equation (ii), we have, N N N [ f N ] f N...(iii) Substituting for N in equation (ii) from equation (iii), we have, N...(iv) f we define the anticock wise sense of rotation as the +ve direction of rotation, then, the cock wise sense becomes the ve direction for the same. Hence, anguar acceeration in the present case becomes negative for this choice of refrence direction. The anguar acceeration, torque due to f and f about C moment of inertia about the axis of rotation

11 46 f R f R mr ( N N) mr ( N ) N mr mr g R f the cyinder had the anguar veocity at t = and at some time t it has an anguar veocity, and in this duration it has turned through an ange, then, 4 g R f the cyinder stops having rotated through an ange, then at, =. Therefore, 4 g R R ( ) 4 g( ) Therefore, the number of rotations accompished by the cyinder, before it stops, n R( ) 8 g ( )

Parallel-Axis Theorem

Parallel-Axis Theorem Parae-Axis Theorem In the previous exampes, the axis of rotation coincided with the axis of symmetry of the object For an arbitrary axis, the paraeaxis theorem often simpifies cacuations The theorem states