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Physics 1D03 Version number 0612345 J. P. Student Multiple-choice answer sheets: HB pencil only; ink will not work Fill circle completely No extra marks.

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Presentation on theme: "Physics 1D03 Version number 0612345 J. P. Student Multiple-choice answer sheets: HB pencil only; ink will not work Fill circle completely No extra marks."— Presentation transcript:

1 Physics 1D03 Version number 0612345 J. P. Student Multiple-choice answer sheets: HB pencil only; ink will not work Fill circle completely No extra marks in answer area Erase well to change an answer

2 Physics 1D03 Centre of Mass Serway 12.1-12.2; and parts of 9.5 Centre of Mass, Centre of Gravity

3 Physics 1D03 Example: The uniform beam (weight 400 N, length L) is supported by a pin at one end, and a scale that can be placed anywhere along the beam. What will the scale read if it is scale a)at the end of the beam (shown)? b)at the centre of the beam? c)at distance ¼ L from the pin? d)at distance ¾ L from the pin?

4 Physics 1D03 Where should we choose the “pivot point” when calculating torques? Answer: In Statics problems, it doesn’t matter. Theorem: If the net force on a body is zero, then the total torque due to all forces will not depend on the choice of “pivot point”. In particular, if F net = 0, then if  net = 0 for one “pivot point”,  net is also zero for every other pivot point.

5 Physics 1D03 The centre of mass is an “average position” of the mass. x1x1 x2x2 x CM m1m1 m2m2 The centre of mass (CM) is defined as the location: Centre of Mass, Centre of Gravity Two particles on the x axis; total mass, M = m 1 + m 2.

6 Physics 1D03 (Recall the position vector r has components x, y, z in R 3.) For many particles: Symmetry: The center of mass of any symmetric object lies on the axis of symmetry and on any plane of symmetry. For a uniform object, CM = center of geometry and mass is distributed evenly around this point

7 Physics 1D03 Examples: The CM can be outside the object. (It is on the line between the centres of the two rectangles which make up the “L”.) The CM is on the symmetry axes. m 2m 2L/3L/3 CM The CM is closer to the more massive object

8 Physics 1D03 Example: Three particles m 1 =m 2 =1.0kg and m 3 =2kg are located like so: y(m) x(m) 1 2 1 2 3 m3m3 m1m1 m2m2 What is the center of mass of the system ? r CM

9 Physics 1D03 Quiz: Baseball bat A baseball bat is sawn in half at its centre of mass. Which piece is heavier? A)The short piece B)The long piece C)Both pieces have the same mass.

10 Physics 1D03 For an object, like the baseball bat, torques due to the gravitational forces are zero around the center of mass But, this is not the point around which mass is evenly distributed b/c the object has a non-uniform mass distribution The lighter side of the bat has a larger moment of inertia, since it is longer

11 Physics 1D03 Centre of Gravity The CM is also the location of the centre of gravity. When we consider the rotational equilibrium of a rigid body, we can treat the gravitational force as if it were a single force applied at the centre of mass. A suspended object will hang with its CM vertically below the point of suspension. CM mg

12 Physics 1D03 Quiz: A hemisphere on a ramp. A uniform solid hemisphere is placed on a ramp. Which of the pictures shows how it rests in equilibrium? A)The top picture B)The middle picture C)The bottom picture

13 Physics 1D03 Centre of Gravity Why can we place the gravitational force at the CM? Calculate the torque (about O) due to the three weights on the x axis: m1gm1g m2gm2g m3gm3g O x1x1 x2x2 x3x3 Now calculate the torque as if a single, total weight were placed at the CM: Mg O x CM The two methods give the same torque ! but

14 Physics 1D03 Replacing the real gravitational forces by a single force, equal to the total weight, and placed at the centre of gravity (which may or may not have any mass at this point), will not change the total gravitational torque (about any pivot). In a uniform gravitational field, the centre of gravity is at the centre of mass (eg: a sphere). This means that the external forces needed to hold a rigid body in equilibrium can be calculated as if gravity were a single force applied at the centre of gravity.

15 Physics 1D03 Example: how far can a pile of bricks lean without falling over? Is it possible for the top brick to be entirely past the edge of the table? ?

16 Physics 1D03 Example: how far can a pile of bricks lean without falling over? E If the centre of gravity of the entire pile is past the edge of the table, there will be a clockwise torque about the edge of the table (point E), and the whole stack will tip (rotate clockwise) about E. If the centre of gravity of the pile is not past the edge of the table, the gravitational torque (about E) will be counterclockwise, and the stack will be stable (at least it will not tip at E). The 4th brick can be entirely past the edge

17 Physics 1D03 Summary: Center of mass: Gravitational forces may be replaced by a single force (weight) located at the CM For a uniform object, CM = center of geometry and mass is distributed evenly around this point


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