When the applied force is not perpendicular to the crowbar, for example, the lever arm is found by drawing the perpendicular line from the fulcrum to the line of action of the force. We call torques that produce counterclockwise rotation positive, and torques that produce clockwise rotation negative.
What is the net torque acting on the merrygo-round? a) +36 N m b) -36 N m c) +96 N m d) -60 N m e) +126 N m 96 N m (counterclockwise) - 60 N m (clockwise) = +36 N m (counterclockwise)
How far do we have to place the 3-N weight from the fulcrum to balance the system? a) 2 cm b) 27 cm c) 33 cm d) 53 cm F = 3 N l = / F = +1 N m = (+1 N m) / (3 N) = 0.33 m = 33 cm
The center of gravity is the point about which the weight of the object itself exerts no torque. We can locate the center of gravity by finding the point where it balances on a fulcrum. What s the center of gravity of a disk? A. any point on the edge of the disk. B. Center of the disk C. Any point half way between the center and the edge.
The center of gravity For a more complex object, we locate the center of gravity by suspending the object from two different points, drawing a line straight down from the point of suspension in each case, and locating the point of intersection of the two lines.
1J-21 Center of Gravity of an Irregular Lamina How can we find the Center of Gravity of Irregular shapes? Where is the Center of Gravity of the cut board? Why does the mass on the string hang straight down? 2/21/2012 Physics 214 2/21/2012 Fall 2009 6 6
If the center of gravity lies below the pivot point, the object will automatically regain its balance when disturbed. The center of gravity returns to the position directly below the pivot point, where the weight of the object produces no torque.
1J-23 Corks & Forks Can the Center of Gravity lie at a point not on the object? How difficult is it to balance this system on a sharp point? Where is the C of G? THE CENTER OF GRAVITY IS NOT ON THE OBJECT. IT ACTUALLY LIES ALONG THE VERTICAL BELOW THE SHARP POINT. WHEN THE FORK IS MOVED THE CM RISES AND THIS MEANS THE SYSTEM IS IN STABLE EQUILIBRIUM. 2/21/2012 Physics 214 2/21/2012 Fall 2009 8 8
1J - 24 double cone What happens to the center of mass? A). Going down the hill B). Going uphill C). Stay at rest D). Depend on the object shape 2/21/2012 9
1J-16 Walk the Plank What happens when a mass is placed at the end of a massive plank? Sum of Torque about Pivot X M g x m g = 0 m = M X / x Can you safely walk to the end of the plank? One can solve for either M or m, if the other quantity is known EVEN WITH A MASS AT THE END OF THE PLANK, THE SYSTEM CAN STILL BE IN EQUILIBRIUM 2/21/2012 Physics 214 2/21/2012 Fall 2009 10 10
1J-28 Wine Bottle Holder Balance a Bottle and a Wooden Holder by Eliminating Net Torque M How does this system Balance? m 1 x1 x m2 2 Sum of Torque about Pivot m 1 x 1 g - m 2 x 2 g = 0 THE CENTER OF GRAVITY OF THE BOTTLE PLUS THE WOOD MUST LIE DIRECTLY OVER AND WITHIN THE BOUNDARY OF THE SUPPORT (PIVOT). FOR BALANCE THERE CAN BE NO NET TORQUE ON SYSTEM. 2/21/2012 Physics 214 2/21/2012 Fall 2009 11 11
How far can the child walk without tipping the plank? For a uniform plank, its center of gravity is at its geometric center. The pivot point will be the edge of the supporting platform. The plank will not tip as long as the counterclockwise torque from the weight of the plank is larger than the clockwise torque from the weight of the child. The plank will verge on tipping when the magnitude of the torque of the child equals that of the plank.
An 80-N plank is placed on a dock as shown. The plank is uniform in density so the center of gravity of the plank is located at the center of the plank. A 150-N boy standing on the plank walks out slowly from the edge of the dock. What is the torque exerted by the weight of the plank about the pivot point at the edge of the dock? a) +80 N m b) -80 N m c) +160 N m d) -160 N m e) +240 N m f) -240 N m
Rotational Inertia and Newton s Second Law In linear motion, net force and mass determine the acceleration of an object. For rotational motion, torque determines the rotational acceleration. The rotational counterpart to mass is rotational inertia or moment of inertia. Just as mass represents the resistance to a change in linear motion, rotational inertia is the resistance of an object to change in its rotational motion. Rotational inertia is related to the mass of the object. It also depends on how the mass is distributed about the axis of rotation.
Simplest example: a mass at the end of a light rod A force is applied to the mass in a direction perpendicular to the rod. The rod and mass will begin to rotate about the fixed axis at the other end of the rod. The farther the mass is from the axis, the faster it moves for a given rotational velocity.
Simplest example: a mass at the end of a light rod To produce the same rotational acceleration, a mass at the end of the rod must receive a larger linear acceleration than one nearer the axis. F = ma It is harder to get the system rotating when the mass is at the end of the rod than when it is nearer to the axis. I case the distance are equal, it s harder to move a heavier mass.
Rotational Inertia and Newton s Second Law Newton s second law for linear motion: F net = ma Newton s second law for rotational motion: F R = m R v = ω R F R = m R net = I The rotational acceleration produced is equal to the torque divided by the rotational inertia.