Class Notes (834,991)
Physics (664)
PHYSICS 1L03 (115)
Reza Nejat (43)
Lecture

# Set7-Ave-1.pdf

4 Pages
213 Views

School
Department
Physics
Course
PHYSICS 1L03
Professor
Reza Nejat
Semester
Fall

Description
Phys 1L03 Tutorial session Set 7 Seventh series: Questions on Center of mass, and Torque Concept: 1. Three masses, 2.0 kg, 3.0 kg, and 6.0 kg, are located at points (3.0, 0), (6.0, 0), and (-4.0, 0), respectively, in meters from the origin. Where is the center of mass of this system? 2. A dumbbell, Figure below, has a connecting bar of negligible mass. Find the location of the center of mass a. if m 1nd m ar2 each 5.0 kg and b. if m1 is.0 kg and m i2 10.0 kg. 3. A water molecule consists of an oxygen atom and two hydrogen atoms. Oxygen has a mass of 16 atomic units (u) and each hydrogen has a mass of 1 u. The hydrogen atoms are each at an average distance of 96 pm ( 96 x 10 -12 m) from the oxygen atom, and are separated from one another by an angle of 104.5 . Find the center of mass of a water molecule. 4. Two children are balanced on a see-saw of negligible mass. (This assumption is made to keep the example simple.) The first child, sitting on left side of the pivot, has a mass of 26.0 kg and sits 1.60 m from the pivot. The second child, sitting on right side of pivot, has a mass of 32.0 kg and 1.3 m from the pivot. Calculate the position of center of mass of the two children: a. from the first child b. from the pivot point. 5. Three balls A, B, and C, with masses of 3 kg, 1 kg, and 1 kg, respectively, are connected by massless rods. The balls are located as in Figure below. What are the coordinates of the center of mass? 6. Alley club-ax consists of a symmetrical 8 kg stone attached to the end of a uniform 2.5 kg stick that is 98 cm long. The dimensions of the club-ax are shown in Figure blow. How far is the center of mass from the handle end of the club-ax? Phys 1L03 Tutorial session Set 7 Seventh series: Questions on Torque, and Levers Concept: Torque : is a tendency of a force to change the rotational motion. Torque [τ] = (Force applied [F]) x (lever arm [r ⊥) Where: the lever arm, r⊥,is defined as the perpendicular distance from the axis of rotation to the line of action of the force. Or: τ = F r sinθ τ = F r ⊥ τ = F r ⊥ Where: r ⊥is the lever arm and ⊥ isthe force component perpendicular to the line of action of the force. Lever: is a rigid bar (e.g. Bone) tha
More Less

Related notes for PHYSICS 1L03
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.