Center of Mass / Torque

Transcription

1 Center of Mass / Torque Level : Physics I Teacher : Kim 1. Center of Mass( or Center of Gravity or Center of Weight) - The center of mass(com) of an object is the point located at the object s average position of weight(or mass). - For a symmetrical object, COM is at the geometrical center of the object. ex) ball, donut,... - For a irregularly shaped object, which can have more weight(mass) at one end than another part of the object, the COM can be located where there is more weight(mass) ex) baseball bat, wrench,.. - The COM does not have to be located where there is mass ex) donut, boomerang,.. - The center of mass can be thought of as a single point located in the object, where all the mass is focused at that single point and gravitational force is pulling at that single point. - If you throw an object in the air with a spin, all objects will naturally spin around its center of a mass. - Even if the object is not symmetrical, the motion of the object will be wobbly while in flight, but the center of mass moves as a single point that follows a smooth projectile motion. - See Video Trajectory of the Center of Mass from MIT Physics department What is useful about Center of Mass - If you spin any shaped object, it will always spin around its center of mass. That is, the center of mass become the axis of the rotating object. - If you throw an irregularly shaped object in the air, the motion will be very wobbly and messy. - However, if we just focus on the center of mass, or point mass, the object will display a smoothed projectile motion. This will simply the motion. - London buses and race cars have very low center of mass, so they not will tilt over.

2 How to find the Center of Mass We can think that all the mass is focused at the center of mass and gravity pulling on that single point. So if we balance any object with one finger, then that point is the center of mass. Answer the following questions. Explain in details 1. If two identical spherical objects were connected to a massless thin rod and spun on top of a frictionless surface, which is most likely point it will spin around? A, B, C? A B C 2. If one spherical object is larger than the other, which is most likely point it will spin around? A, B, C? A B C 3. If a star wobbles off its center of mass, what does that suggest? 4. Explain which of following structure will topple. (a) (b)

3 5. Why do you spread your feet farther apart when standing in a bumpy-riding bus? Demo) Balancing a soda can 6. Why does leaning Tower of Pisa not toppled? 7. What is the role of the heavy tail of a dinosaur? 8. Why does a ball roll down a slope?

4 9. Why can you not successfully bend over and touch your toes when you stand your heels against the wall? 1.1 Finding the Center of Mass in 1-dimension - The COM of two particles is located on the x axis and lies somewhere between the particles. Y x1 m1 xcm x2 CM m2 X x cm = m 1x 1 +m 2 x 2 m 1 +m 2 - The COM for a system of many particles in the x-coordinate x cm = m 1x 1 +m 2 x 2 +m 3 x 3 + m n x n m 1 +m 2 +m 3 + m n = m ix i m i Q1) Find the COM of m 1=2kg, m 2=2kg where the distance from each other is 4m. x1=0m x2=4m Q2) Find the COM of m 1=2kg, m 2=4kg where x 1 = 2m, x 2 = 6m. 0 x1=2m x2=6m Q3) Four objects are situated along the x-axis. A 2kg object is at +3m, a 3kg is at +2.5m, a 2.5kg object is at the origin, and a 4kg object is at. Where is the center of mass of these objects? 4kg 2.5kg 3kg 2kg x-axis(m)

5 2. Torque Torque is the measure of the ability of a force to rotate an object around an axis. Torque is defined as τ = rfsinφ [N.m] where τ is the torque, r is the distance between the axis and the position of the force applied and φ is the angle between the force and lever arm. Axis r F φ lever arm - Torque will be maximum when φ=90. That is rotation of the object will occur best(easier) when a force is applied φ=90 - This is the reason why the doorknob is position as far away from the axis (hinge of the door) as possible. - We choose counter-clockwise as positive direction *~To unscrew a bolt, we need to rotate the bolt. Whenever you want to rotate any object, you must apply a torque on the object. Q4) Calculate the torque produced by a 50N perpendicular force at the end of a 0.08m long wrench. Use τ = r F a) 4N m b) 12N m c) 25N m d) 35N m 0.08m Push with 50N of force! Q5) Calculate the torque produced by a 50N perpendicular force at the end of a 0.1m long wrench. Use τ = r F a) 10N m b) 5N m c) 2N m d) 0.5N m 0.1m Push with 50N of force! Q6) To unscrew a bolt, you need to apply a torque. Let s say a bolt needs 42N.m of torque to be unscrewed. i) If the length of the wrench you have is 0.15m and the maximum strength you can apply on the handle is 200N, will the bolt become loosened? If not, you need to get a longer wrench. How long must be the wrench in order to loosen the bolt?

6 2.1 Net Torque There can be more than one torque acting on an object. (compare net force=> F=ma) where τ = r Fsinφ τ =τ 1 + τ 2 + τ n Counter-clockwise is positive! Q7) A meter-stick is pivoted at the center. Any unbalance torque applied will result in rotating the meter-stick. If 10N of pushing force are applied as shown at the edge at both ends, find the net torque acting on the system. (and which direction? clockwise or counter-clockwise?) a) 10N m b) 15N m c) 20N m d) 25N m *~1m =100cm axis Q8) A meter-stick is pivoted at the center. Any unbalance torque applied will result in rotating the meterstick. If 15N and 10N of pushing force are applied as shown at the edge at both ends, find the net torque acting on the system. (and which direction? clockwise or counter-clockwise?) a) 1.0N m b) 1.5N m c) 2.0N m d) 2.5N m F=15N Q9) A meter-stick is pivoted at the center. Any unbalance torque applied will result in rotating the meter-stick. If 10N of pushing force are applied as shown, find the net torque acting on the system. (and which direction? clockwise or counter-clockwise?) a) 10N m b) 8N m c) 6N m d) 4N m 0.3m Q10) A meter-stick is pivoted at the center. Any unbalance torque applied will result in rotating the meterstick. If 15N and 10N of pushing force are applied as shown, find the net torque acting on the system. (and which direction? clockwise or counter-clockwise?) a) 1.0N m b) 1.5N m c) 2.0N m d) 2.5N m F=15N 0.2m

7 2.2 Balance Net Torque i) If the stick does not move, the system is in equilibrium. This can only be possible if the net torque is zero. That is, the torque produce on the left of the axis and the right of axis must be equal but opposite direction. (*~remember that counter-clockwise is positive) = 1 + (- 2) = 0 => 1 = 2 ii) If you balance a meter-stick on a fulcrum and hang objects on both sides, the weight force F g of the object will produce net torque. (F g=mg) If you shift the position of the two objects, until the meter-stick balance, the net torque will be zero. By using this theory, we can measure the mass of the unknown objects by hanging known objects until they balance F1 r1 axis axis r2 F2 left = right => r left m left g = r right m right g Example) A meter-stick is placed on a fulcrum at the 50cm mark. If the mass m is placed so that the whole system is in equilibrium, then the torque on the left and the torque on the right is equal. Find the unknown mass. 1meter=100cm, g=9.8m/s 2 10cm 50cm 80cm τleft τright 2kg m=? Sol) A torque on the left of the axis exists due to the gravitational force F g pulling on the 2kg object over a distance of 0.4m. A torque on the right of the axis exists due to the gravitational force F g pulling on the block of unknown mass over a distance of 0.3m. If the block is placed so that the system is balanced, then left = right. Since torque is defined as τ = r Fsinφ and the force F in this system is F g=mg, Solving for m gives m=2.67kg r left m leftg = r right m right g => 0.4 (2 9.8) = 0.3 (m 9.8)

8 left = right => r left m left g = r right m right g Q11) A meter-stick is placed on a fulcrum at the 50cm mark. The system is in equilibrium. Find the unknown mass. r must be in meters 1meter=100cm, g=9.8m/s 2 10cm 30cm 50cm 80cm 2kg 1kg m=? 2.3 Find the mass of Meter-stick If the fulcrum is no longer placed under the center of mass of the meter-stick, then gravitational force F g will pull on the location of center of mass as if all the mass is focused on that single point Q12) A meter-stick is placed on a fulcrum at the 20cm mark. If the mass of 1.2kg is placed at the 0cm so that the whole system is in equilibrium, then the torque on the left and the torque on the right is equal. Find the mass of the meter-stick. 1meter=100cm, g=9.8m/s 2 COM of the meter-stick!! 0cm 20cm 50cm 100cm τleft τright m=1.2kg a) 4.6kg b) 2.4kg c) 1.2kg d) 0.8kg

9 left = right => r left m left g = r right m right g Q13) A meter-stick is placed on a fulcrum at the 20cm mark. Find the mass of the meter-stick COM of the meter-stick!! 10cm 20cm 50cm 0.5kg a) 0.85kg b) 0.36kg c) 0.17kg d) 0.08kg Q14) A meter-stick is placed on a fulcrum at the 30cm mark. The system is in equilibrium. Find the mass of the meter-stick 1meter=100cm, g=9.8m/s 2 COM of the meter-stick!! 10cm 30cm 50cm 80cm 8kg m=2.67kg