Lecture 12! Center of mass! Uniform circular motion!

Transcription

1 Lecture 1 Center of mass Uniform circular motion

2 Today s Topics: Center of mass Uniform circular motion Centripetal acceleration and force Banked cures

3 Define the center of mass The center of mass is a point that represents the aerage location for the total mass of a system. m x x 1 1 cm m m m x For linear motion, we can consider all of the mass of the object to be concentrated at the center of mass.

4 Center of mass simplifies calculations inoling conseration of momentum for combined linear and rotational motion, the CM of an object is the point that obeys the linear equations of motion we hae deeloped DEMO: Shape

5 What if there are more than Oxygen y Oxygen objects? nm 60.0 o Sulfur 60.0 o nm x A sulfur dioxide molecule consists of oxygen atoms and a sulfur atom. The sulfur atom is twice as massie as the oxygen atom. Find the x and y coordinates of the center of mass of this molcule 1. Find the x and y components of all atoms with respect to the origin (sulfur atom) x-coordinate y-coordinate Oxygen atom (on the left) x 1 (0.143 nm) sin nm Oxygen atom (on the right) x +(0.143 nm) sin nm y 1 +(0.143 nm) cos nm y +(0.143 nm) cos nm Sulfur atom x 3 0 nm y 3 0 nm

6 Let m 1 and m represent the masses of the oxygen atoms on the left and right, respectiely. Likewise, let m 3 be the mass of the sulfur atom. The x-coordinate of the center of mass is x cm m x + m x + m x m 1( 0.14 nm) + m nm m 1 + m + m 3 m 1 + m + m 3 Oxygen nm o o ( ) + m 3 ( 0 nm) y Sulfur Oxygen nm x Since m 1 m (both are oxygen atoms), the equation aboe yields x cm 0 m The y-coordinate of the center of mass is y cm m y + m y + m y m 1 ( nm) + m nm m 1 + m + m 3 m 1 + m + m 3 ( ) + m 3 ( 0 nm) Substituting m 1 m m and m 3 m into the equation aboe yields y cm m ( nm) + m ( nm) + m( 0 nm) m + m + m nm

7 Continuous mass distributions Use symmetry CM somewhere along this line The CM need not be inside the object For a system composed of multiple parts with symmetry, first condense each part to its CM and then treat each CM as a pointlike particle kg 1 kg kg 1 kg CM

8 Velocity of the center of mass Δx cm m 1 Δx 1 + m Δx m 1 + m cm m m m 1 + m p 1 + p m 1 + m p total m total In an isolated system, the total linear momentum does not change, therefore the elocity of the center of mass does not change.

9 A stationary 4-kg shell explodes into three pieces. Two of the fragments hae a mass of 1 kg each and moe along the paths shown with a speed of 10 m/s. The third fragment moes upward as shown. kg What is the speed of the third fragment? Momentum along x and y must be consered,so (kg)( y y 5 m/s ) (1kg)(10m/s)(cos60 ) kg 1 kg 10 m/s 10 m/s What is the speed of the center of mass of this system after the explosion? (a) zero m/s (b) 1 m/s (c) 3 m/s (d) 5 m/s (e)7 m/s

10 Uniform Circular Motion Uniform circular motion is the motion of an object traeling at a constant speed on a circular path. Let T be the time it takes for the object to trael once around the circle. π T r T is also known as the period of the motion

11 Example A ball moes with a constant speed of 4 m/s around a circle of radius 0.5 m. What is the period of the motion? T πr T πr π (0.5 m) 4 m/s 0.4 s

12 Circular Motion In uniform circular motion, the speed is constant, but the direction of the elocity ector is not constant. So een if the speed is constant, there has to be an acceleration. centripetal acceleration What is the direction and magnitude of this centripetal acceleration?

13 Centripetal acceleration 0 Δ 0 F a Δ c Δt Δ F 0 a c r The centripetal acceleration in uniform circular motion points to the center of the circle. a c r rˆ ˆr is a unit ector that points radially away from the origin

14 Example 1 A racecar is traeling at constant speed around a circular track. What happens to the centripetal acceleration of the car if the speed is doubled? (a) The centripetal acceleration remains the same. (b) The centripetal acceleration increases by a factor of. (c) The centripetal acceleration increases by a factor of 4. (d) The centripetal acceleration is decreased by a factor of one-half. (e) The centripetal acceleration is decreased by a factor of one-fourth a c r rˆ If 0, 4 0

15 Centripetal Force F a m The centripetal force is the net force required to keep an object moing on a circular path. The direction of the centripetal force always points toward the center of the circle and continually changes direction as the object moes. F c ma c m r rˆ

16 What s the origin of the Centripetal Force? A 0.5-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball reoles twice eery second, what is the tension in the string? Here I am calculating the magnitude of F C F F T T πr T F c m r mac m r π (0.5 m) 0.5 s (0.5 kg) 6.8 m/s ( 6.8 m/s) 0.5 m 0 N In this case, it s the tension in the string

17 How about for a car going around a turn? On an unbanked cure, the static frictional force proides the centripetal force. Looking from behind f S F N mg

18 Or for a car going around a banked turn? On a frictionless banked cure, the centripetal force is the horizontal component of the normal force. The ertical component of the normal force balances the car s weight.

19 F c F sin θ N F cosθ mg N m r tan θ rg

20 An Amusement Park Ride A small child rides in a barrel of fun ride. The floor drops downward and the child remains pinned against the wall. If the radius of the deice is.15 m and the coefficient of static friction between the child and the wall is 0.400, with what minimum speed is the child moing if he is to remain pinned against the wall? F N f S mg F N m r f S max mg 0 µ S F N mg 0 µ S m r mg 0 rg µ S 7.6 m/s Notice does not depend on the mass of the child.