Torque and Center of Mass Handout HW #6. Center of Mass: The center of mass (or mass center) is the mean location of all the mass in a system. The motion.

Torque and Center of Mass

Handout

HW #6

Center of Mass:

The center of mass (or mass center) is the mean location of all the mass in a system.

The motion of an object can be characterized by this point in space. All the mass of the object can be thought of being concentrated at this location. The motion of this point matches the motion of a point particle.

Finding the Center of Mass:

Uniform geometric figures have the center of mass located at the geometric center of the object.

Note that the center of mass does not have to be contained inside the volume of the object.

Collections of Point Masses:

The center of mass for a collection of point masses is the weighted average of the position of the objects in space.

Each object will have a position in space. The center of mass is found as:

321

332211

mmm

xmxmxmxcm

321

332211

mmm

ymymymycm

Example #1: A 10.0 kg mass sits at the origin, and a 30.0 kg mass rests at the 12.0 m mark on the x – axis. (a) Find the center of mass for this system.

01 x

kgm 0.101

mx 0.122

kgm 0.302

321

332211

mmm

xmxmxmxcm

kgkg

mkgkgxcm 0.300.10

0.120.3000.10

m00.9

(b) Find the center of mass for this system relative to the mass at the right.

mx 0.121

kgm 0.101

02 x

kgm 0.302

321

332211

mmm

xmxmxmxcm

kgkg

kgmkgxcm 0.300.10

00.300.120.10

m00.3

Although numerically different, it is the same point in space relative to the masses…

Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below. Determine the center of mass for this object.

Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below. Determine the center of mass for this object.

Treat as two objects:

6 cm object:

gcmcm

gm 40.200.6

0.10

00.41

0,1 cmx

cmy cm 00.3,1

4 cm object:

gcmcm

gm 60.100.4

0.10

00.42

cmx cm 00.2,2 0,2 cmy

321

332211

mmm

xmxmxmxcm

g

cmggxcm 00.4

00.260.1040.2 cm800.0

321

332211

mmm

ymymymycm

g

gcmgycm 00.4

060.100.340.2 cm80.1

Example #3: Determine the center of mass of the following masses, as measured from the left end. Assume the blocks are of the same density.

Mm 1

21olx

Mm 82

olx 22 Mm 273 olx 5.43

321

332211

mmm

xmxmxmxcm

M

lMlMl

Mx

ooo

cm 36

5.427282

ocm lx6

23

Torque

Torque is the rotational equivalent of force. A torque is the result of a force applied to an object that tries to make the object rotate about some pivot point.

Equation of Torque:

pivot point

r distance from pivot to applied force

F

applied force

angle between direction of force and pivot distance.

sinrFtorque

Note that torque is maximum when the angle is 90º.

The units of torque are Nm or newton · meter

The torque is also the product of the distance from the pivot times the component of the force perpendicular to the distance from the pivot.

rFrFtorque sin

The torque is also the product of the force times the lever arm distance, d.

FdrFtorque sin

Example #4: Calculate the torque for the force shown below.

sinrF 0.60sin30000.2 Nm

Nm520

Example #5: Calculate the total torque about point O on the figure below. Take counterclockwise torques to be positive, and clockwise torques to be negative.

sinrF

20sin100.460sin250.2 NmNmnet

Nmnet 6.29

Example #6: The forces applied to the cylinder below are F1 = 6.0 N, F2 = 4.0 N, F3 = 2.0 N, and F4 = 5.0 N. Also, R1 = 5.0 cm and R2 = 12 cm. Determine the net torque on the cylinder.

90sin21 RFnet

90sin22 RF

90sin13 RF

0sin24 RF0

mNmNmNnet 050.00.212.00.412.00.6

Nmnet 14.0