Question

A homogeneous rod of length L=1.47 mL=1.47 m and mass m=4.43 kgm=4.43 kg is pivoted about...

A homogeneous rod of length L=1.47 mL=1.47 m and mass m=4.43 kgm=4.43 kg is pivoted about one end. It starts at an angle of θ=27.8 ∘θ=27.8 ∘ with the vertical as shown in Figure 1. You may find it useful to use I=13mL2I=13mL2 for a rod pivoted about one end. The rod has just been released from rest at θ.

1) What is the magnitude of the torque acting on the rod in this position?

2) What is the angular acceleration of the rod?

3) What is the acceleration of a point on the end of the rod?

Science   Physics   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Torque is product of perpendicular distance of the force form the pivot and the force.

By the relation of second law of circular motion, we can calculate angular acceleration.

And the resultant acceleration of the point at the end of the rod is the resultant of tangential and angular acceleration.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
A uniform rod of mass M and length L is pivoted at one end. The rod...
A uniform rod of mass M and length L is pivoted at one end. The rod is left to freely rotate under the influence of its own weight. Find its angular acceleration α when it makes an angle 30° with the vertical axis. Solve for M=1 Kg, L=1 m, take g=10 m s-2. Hint: Find the center of mass for the rod, and calculate the torque, then apply Newton as τ= Ι·α 
A uniform rod of mass M and length L is pivoted at one end. The rod...
A uniform rod of mass M and length L is pivoted at one end. The rod is left to freely rotate under the influence of its own weight. Find its angular acceleration α when it makes an angle 30° with the vertical axis. Solve for M=1 Kg, L=1 m, take g=10 m s-2. Your answer in X.X rad s-2. Hint: Find the center of mass for the rod, and calculate the torque, then apply Newton as τ= Ι·α
A 150 g uniform rod of length 3.0 m is pivoted about one end. What would...
A 150 g uniform rod of length 3.0 m is pivoted about one end. What would be the angular acceleration due to gravity if the rod were held horizontally and released from rest?
A stick of mass M and length L is pivoted at one end. A small mass...
A stick of mass M and length L is pivoted at one end. A small mass m<M is attached to the right-hand end of the stick. The stick is held horizontally and released from rest. Given that the rotational inertia of a uniform rod pivoted around one end is 1/3ML^2, determine the rotational inertia of the described system. Calculate the angular velocity of the system when it reaches a vertical position. You cannot use rotational kinematics here because angular acceleration...
The uniform thin rod in the figure below has mass M = 2.00 kg and length...
The uniform thin rod in the figure below has mass M = 2.00 kg and length L = 2.87 m and is free to rotate on a frictionless pin. At the instant the rod is released from rest in the horizontal position, find the magnitude of the rod's angular acceleration, the tangential acceleration of the rod's center of mass, and the tangential acceleration of the rod's free end. HINT An illustration shows the horizontal initial position and vertical final position...
A uniform thin rod of length 0.45 m and mass 6.5 kg can rotate in a...
A uniform thin rod of length 0.45 m and mass 6.5 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet velocity makes an angle of 60° with the rod (see the figure). If the bullet lodges in the rod and the...
A pendulum rod has a length L and has a mass m at the end. The...
A pendulum rod has a length L and has a mass m at the end. The pendulum is released from rest at an angle θ away from straight down. (In this problem you may neglect friction and mass of the pendulum arm.) What will the velocity of the mass be at the bottom? What will the tension in the pendulum arm be as it rotates through the bottom of its swing? A person holds the pendulum in place before it...
A uniform thin rod of length 0.56 m and mass 3.2 kg can rotate in a...
A uniform thin rod of length 0.56 m and mass 3.2 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a 3.5 g bullet traveling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet's path makes angle θ = 60° with the rod. If the bullet lodges in the rod and the angular velocity of the rod is 12.0 rad/s...
A rod with a length of 80 cm is pivoted about a horizontal axis through its...
A rod with a length of 80 cm is pivoted about a horizontal axis through its center (rod's mass is negligable). A black mouse with a mass of  0.510 kg clings to one end of the stick, and a brown rat with a mass of  0.211 kg clings to the other end. The system is released from rest with the rod horizontal. Part A If black mouse and brown rat can hold on, Find their velocities as the rod swings through a...
A thin rod of length L=2.0 m can rotate freely about a pin that is located...
A thin rod of length L=2.0 m can rotate freely about a pin that is located a distance x=0.44 m from one end. Two forces act on opposite ends of the rod. Force F1 =10.2 N pushes at an angle θ1 =119.3° with respect to the rod (a distance x from the pivot point). Force F2 =5.0 N pushes on the opposite end at an angle θ2 =68.5° with respect to the rod as shown in the figure. Find the...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT