Question

Proof center of gravity of a rod by length of L is: Xcom= 1/2L.

Proof center of gravity of a rod by length of L is: Xcom= 1/2L.

0 0
Add a comment Improve this question Transcribed image text
Know the answer?
Add Answer to:
Proof center of gravity of a rod by length of L is: Xcom= 1/2L.
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 1. Locate the center of gravity of the homogeneous rod bent in the form of a...

    1. Locate the center of gravity of the homogeneous rod bent in the form of a semicircular arc. The rod has a weight per unit length of 3 N/m. Also, determine the horizontal reaction at the smooth support B and the x and y components of reaction at the pin A.

  • AXis Two spheres of mass m are attached to a light rod of length 2L. The...

    AXis Two spheres of mass m are attached to a light rod of length 2L. The rod is fixed at its center to a freely-rotating axis. The rod is horizontal when a small bird of mass 5m carefully lands on the right sphere. Express a answers in terms of m, L, and g (a) What is the net torque on the system the instant the bird lands (while the rod is still horizontal)? (b) Determine the angular acceleration of the...

  • (Figure 1) A thin rod of mass mr and length 2L is allowed to pivot freely about its center, as shown in the diagram.

    Pivoted Rod with Unequal Masses (Figure 1) A thin rod of mass mr and length 2L is allowed to pivot freely about its center, as shown in the diagram. A small sphere of mass m1 is attached to the left end of the rod, and a small sphere of mass m2 is attached to the right end. The spheres are small enough that they can be considered point particles. The gravitational force acts downward, with the magnitude of the gravitational acceleration...

  • Problem-2 A very thin, uniformly-charged rod with total charge Q and length 2L lies along the...

    Problem-2 A very thin, uniformly-charged rod with total charge Q and length 2L lies along the vertical 2-axis. Sitting still (zero velocity and zero acceleration) at a distance H above the top end of the rod is a proton. Gravity is acting on the proton in the downward (-2) direction. What is the the total charge Q on the rod? + profon K & de d = k l . dx. - EL - K Let =) , U, -...

  • 1. A uniform rod of mass M = 5.01kg and length L = 1.18m can pivot...

    1. A uniform rod of mass M = 5.01kg and length L = 1.18m can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as seen in the figure below. 2. Determine the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of the rod, as shown. Please show work for both questions radusn2. The rod is held horizontally and then released. At the moment...

  • a) Calculate the electrostatic force on a uniformly charged rod of length 2l,and charge q which...

    a) Calculate the electrostatic force on a uniformly charged rod of length 2l,and charge q which lies along the axis of a uniformly charged ring of radius R and charge q'. The centers of the charged rod and the rings are displaced by z = z0 . (b) Show that if z0 >>l,is satisfied, then the expression of calculated force reduces to that between point charges.

  • 3) A U-shaped rod has vertical sides of length L and a horizontal base of length...

    3) A U-shaped rod has vertical sides of length L and a horizontal base of length 2L as shown below. The vertical sides each have a charge per unit length of +A while the base has a charge per unit length of -. Choose a cartesian coordinate system with its origin at the center of the base and with the x- and y- axes horizontal and vertical, respectively a. Solve for the approximate electric potential at the point (0,y) where...

  • A thin rod of mass M and length L has a fixed rotation axis a distance...

    A thin rod of mass M and length L has a fixed rotation axis a distance L/6 from one end. (a) Using the parallel-axis theorem, find the moment of inertia of the rod about its rotation axis. (b) Suppose the rod is held horizontally at rest and then released. Draw a free-body diagram of the rod at the moment of its release, and find its angular acceleration at this moment. (Remember that gravity acts at the rod’s center.) (c) Find...

  • A uniform rod of mass M = 5.14kg and length L = 1.01m can pivot freely...

    A uniform rod of mass M = 5.14kg and length L = 1.01m can pivot freely (i.e., we ignore friction) about a hinge attached to a wall, as seen in the figure below. The rod is held horizontally and then released. At the moment of release, determine the angular acceleration of the rod. Use units of rad/s^2. Determine the linear acceleration of the tip of the rod. Assume that the force of gravity acts at the center of mass of...

  • A rod, in the form of a cylinder of length L and circular cross-section of radius...

    A rod, in the form of a cylinder of length L and circular cross-section of radius r, has a resistance R. What is the new value of the resistance R if the length is doubled (2L) and the radius is doubled (2r)?

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT