rolling_without_slip

# rolling_without_slip - Rolling Without Slip; etc A...

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Unformatted text preview: Rolling Without Slip; etc A particularly interesting case of rotation arises when a ring or disk or cylinder or sphere rolls along a solid surface. Case I: Rolling without slipping ' Consider the case where surface is horizontal and the roller has a constant velocity VC = VC 2? at its center T CP 11 7.: VC is constant so acceleration g = 0. No force involved. If there is no SLIP, the velocity at the point of contact P must be ZERO at ALL times. That is, the point on the circle Which comes into contact with the surface changes with time but at the instant of contact VP = 0 always. To achieve this, the object must have an angular velocity g3 such that the tangential velocity V, at P, due to the rotation, is exactly equal and opposite to VC. -—r This will ensure that VP =_VC +V, = a, = _C - R and for the case shown in the ﬁgure (0 = —— KC— 2 H R Qis constant sog = 0[N0 TORQUE] It is interesting to ask what are the velocities at the points A, C, B and T in the roller. R R0) V: AC=— V =VV ————“ =—C£ (_ 2) —"» —C> 2 x 2 VC:VC R Rd) A 3 A (BC=—2—) =E———x=§ch Case 11: without slipping. Let us put our roller on an inclined plane and let it roll down the incline b Now it will have both a linear acceleration and an angular acceleration. We have drawn all the effective forces acting on the roller. For the linear acceleration (MQ=Z§) —Ma=—MgSz'nt9+fS —>(l) For the angular acceleration Ga=23) —Ia=~Rfs —>(2) Since there is no slip, velocity and acceleration at P must be ZERO at all times and this requires From (2) and (3) and substituting in (1) Moments of Inertia Ring I = M R2 2 Disk I = M R 2 2 Cylinder 1 = M R 2 Sphere (hollow)rI =§M R2 Sphere (solid) 1 =§~MR2 a:E —>(3) Ia Ia fs =T=7€i Ma=MgSin6—%E:— a= “’36 ' —>(4) 1+ MR2 Hence a is independent of M and R. It only depends on how mass is distributed around the axis of rotation. Clearly, the ring has the smallest acceleration — g SinH ﬁring : 2 5e and the solid sphere has the largest acceleration — g 81716 A a = x —v SJ. 1 '4 Next, it must be realized that the static friction force cannot axceed ,u_v 11 Because f, S ,us n So f, s ,u, Mg COSQ From Eq (1) and Eq (4) f, = MgSin<9 — Ma =MgSin61—wl—I— 1+ 2 MR I ' M R2 I M R2- So if we start increasing 0 eventually fs becomes equal to its largest value and the roller will slip = MgSinQ 1+ I c 2 . MgSinQ —% =ysMgC056 1+ 2 MR . 1+ I 2 tanH = MR #3 1 MR2 Ring will be the ﬁrst to slip [tan 9 = 2a,] Note In the above motion, the force of gravity provided the linear acceleration and fs provided the torque. Case III: It is interesting to compare this with the way your automobile gets going on a horizontal surface. The tires are fairly complex but we will treat them as rigidplgodies (rings). We need static friction (as anyone who has tried to get going on an icy road knows, the tires spin in place). But now'the Torgue is provided by the engine (as you engage the gear) and the tire pushes back on the road with fS and 'by Newton’s Third Law the road pushes the car forward. Again ﬂsmn ' m=M9 J fr S as Mg and as'always Mg = E = fs so > . a S ,usg Maximum acceleration is ,us g in magnitude. Case IV: After comparing case I and case IV you can begin to understand why while driving on a slippery road it is recommended that one maintains a constant speed (E = 0) and deﬁnitely must avoid excessive use of acceleration/ brake ( E i 0). Case V: When you go bowling you throw the ball so that when it arrives on the Shute surface it has a linear velocity Vi 9i: and it slips along the surface. However, once it touches the surface kinetic friction comes into play. Let us see how this leads to rolling, without slip. We will take the {ﬁe-i; general case of the roller being sphere, ring, and cylinder. ‘ [lg = *ﬂkMng [n~Mg = 0] so, g=—,ukgfc V and y=(v,—,ukgt)£ However, now there is also a torque about the axis through the center E = "R f k 2 so there is an angular acceleration _R " [19. = I] a = If" Z where I is the moment of Inertia. The angular velocity w = O — R fkti H I w = _ ,ukMgR tz "’ I and to get the condition for case I [w = , we can look for time t, when M R21: vl_lukgtl:+,uk g 1 I _ Vi t' _ M R2 :“kg 1+ I Notice t1, is also independent of M and R since I = (const) x MR2 for all rollers. At later times we have pure roll, 3 = const., 143 = const. and there is no force or torque on the roller (case I). ’ ‘ At later times we have pure roll, 12 = const., Lv = const. and there is no force or torque on HoWever, now there is also a torque about the axis through the center 3 = —R fk 2 so there is an angular acceleration . _ R A [126. = z] a = If" 2 where I is the moment of Inertia. I The angular velocity w : O _ R 'fkt z _' I w = _ ,ukMg R t z a I and to get the condition for case I [w = , we can look for time t; when M R2 vi — :“kg t1 = + I _ Vi t‘ — MR2 Iukg 1+ 1 Notice t1, is also independent of M and R since I = (const) x MR2 for all rollers. the roller (case I). ...
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## rolling_without_slip - Rolling Without Slip; etc A...

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