ch11 kho tài liệu bách khoa
➤ Gửi thông báo lỗi ⚠️ Báo cáo tài liệu vi phạmNội dung chi tiết: ch11 kho tài liệu bách khoa
ch11 kho tài liệu bách khoa
https://khothuvien.cori!I.The initial speed of the car is V = (80.0X 1000/3600) = 22.2 m/s. The tire radius is A’ = 0.750/2 = 0.375 in.(a)The initial ch11 kho tài liệu bách khoa speed of the car is the initial speed of the center of mass of the tire, so Eq. 11 -2 leads to1'^1 R22.20.375= 59.3rad/s.(b)With 0= (3O.O)(2n) = 188 rad and (!)— 0. Eq. 10-14 leads to(ớ2 = ứ)20 + lad => |ứ| =59.3*2(188)= 9.31 rad/s2.(c)Eq. 11-1 gives RO- 70.7 m for the distance traveled.https://kho ch11 kho tài liệu bách khoathuvien.cori!2The velocity of the car is a constant v = +(80) (1000/3600) = (+2 2 m/s)i, and the radius of the wheel is r = 0.66/2 = 0.33 in.(a)In thech11 kho tài liệu bách khoa
car’s reference frame (where the lady perceives herself to be at rest) the road is moving towards the rear at rr<1Mj = -v = -22m/s, and the motion ofhttps://khothuvien.cori!I.The initial speed of the car is V = (80.0X 1000/3600) = 22.2 m/s. The tire radius is A’ = 0.750/2 = 0.375 in.(a)The initial ch11 kho tài liệu bách khoaslational) in this frame. Eq. 10-18 gives =(+22m/s)i.(c)The bottom-most point of the tire is (momentarily) in firm contact with the road (notskidding) and has the same velocity as the road:= (-22m/s)i. This also followsfrom Eq. 10-18.(d)This frame of reference is not accelerating, so “fixed” points ch11 kho tài liệu bách khoawithin it have zero acceleration; thus, Occntcr = 0.(e)Not only is the motion purely rotational in this frame, but we also have to- constant, which mech11 kho tài liệu bách khoa
ans the only acceleration for points on the rim is radial (centripetal). Therefore, the magnitude of the acceleration isV2 222. ya = — == 1.5 X 10 m/shttps://khothuvien.cori!I.The initial speed of the car is V = (80.0X 1000/3600) = 22.2 m/s. The tire radius is A’ = 0.750/2 = 0.375 in.(a)The initial ch11 kho tài liệu bách khoae of reference (where the road is “fixed" and it is the car that appears IO be moving). The center of the tire undergoes purely translational motion while points al the rim undergo a combination of translational and rotational motions. The velocity of the center of the tire is V = (+22 tn/s)i.(h)In ch11 kho tài liệu bách khoapan (b). we found vutUBf = +v and we use Eq. 4-39:I top. Í round I top, car + ^car.trvund V 1 + V 1211w hich yields 2v = +44 m/s. This is consistent wch11 kho tài liệu bách khoa
ith Fig. 11-3(c).(i)We can proceed as in pan (h) or simply recall that the bottom-most point is in firm contact with the (zero-velocity) road. Either https://khothuvien.cori!I.The initial speed of the car is V = (80.0X 1000/3600) = 22.2 m/s. The tire radius is A’ = 0.750/2 = 0.375 in.(a)The initial ch11 kho tài liệu bách khoaelocity frames of reference, the accelerations are unaffected. The answer is as it was in pail (e): 1.5 X 10' m/s2.(T) As explained in pari (k). a = 1.5 X IO3 ni/s2.3By Eq. 10-52. the work required to stop the hoop is the negative of the initial kinetic energy of the hoop. The initial kinetic energy ch11 kho tài liệu bách khoa is K = 4/ftT + |/?JV2 (Eq. 11-5). where / = inR2 is its rotational inertia about the center of mass. III = 140 kg. and V = 0.150 m/s is the speed ofch11 kho tài liệu bách khoa
its center of mass. Eq. 11-2 relates the angular speed to the speed of the center of mass: (i)= v/R. Thus.K =i+^mv2 = MV2 = (I4O)(O.15O)2https://khothuvien.cori!I.The initial speed of the car is V = (80.0X 1000/3600) = 22.2 m/s. The tire radius is A’ = 0.750/2 = 0.375 in.(a)The initial https://khothuvien.cori!I.The initial speed of the car is V = (80.0X 1000/3600) = 22.2 m/s. The tire radius is A’ = 0.750/2 = 0.375 in.(a)The initialGọi ngay
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