A rod of length L = 1.58 m and mass m = 2.02 kg has a constant density and pivots from one end. It begins at an angle of 0 = 22.8 from the vertical as shown in Figure 1. The moment of inertia for a rod pivoting about one end is / = mL?. in the figure the rod has just been released from rest at angle 0. Figure 1: Diagrammatic representation of a rod-shaped pendulum of length L pivoting about one end from a starting angle of 0 degrees to the vertical. Part 1) What is the magnitude of the torque acting on the rod in this position? Nm Part 2) What is the angular acceleration of the rod? a = rad s-2 Part 3) What is the acceleration of a point on the end of the rod? ms-2

A rod of length L = 1.58 m and mass m = 2.02 kg has a constant density and pivots from one end. It begins at an angle of 0 = 22.8 from the vertical as shown in Figure 1. The moment of inertia for a rod pivoting about one end is / = mL?. in the figure the rod has just been released from rest at angle 0. Figure 1: Diagrammatic representation of a rod-shaped pendulum of length L pivoting about one end from a starting angle of 0 degrees to the vertical. Part 1) What is the magnitude of the torque acting on the rod in this position? Nm Part 2) What is the angular acceleration of the rod? a = rad s-2 Part 3) What is the acceleration of a point on the end of the rod? ms-2

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter12: Rotation I: Kinematics And Dynamics

Section: Chapter Questions

Problem 64PQ: A potters wheel rotates with an angular acceleration = 4at3 3bt2, where t is the time in seconds, a...

See similar textbooks

Similar questions

Consider a thin hoop of mass (1.420 ± 0.001) kg and radius (0.250 ± 0.002) m. The moment of inertia for a thin hoop rotating about an axis going through its center is MR2 . Calculate the moment of inertia of this hoop and its uncertainty using error propagation rules.
Assume that the physical pendulum in a grandfather's clock is composed of a uniform rod (mass mrod = 5M, length Lrod = 10R) pivoted at the top, with its lower end connected to a uniform disk (mass mdisk = M, radius rdisk = R). The expressions for the moments of inertia about the centers of mass are: Irod = (1/12) mrod Lrod2 and Idisk = (1/2) mdisk rdisk2. Using the parallel axis theorem for both (rod and disk), find the moment of inertia of the system about the pivot. Report your answers in terms of M and R. A. (1603/6) MR2 B. (1729/6) MR2 C. (259/6) MR2 D. (253/6) MR2
A 20-centimeter pendulum moves according to the equation ? = 0.27 cos(6t), where ? is the angular displacement from the vertical in radians and t is the time in seconds. Determine the maximum angular displacement ?max and the rate of change of ? when t = 5 seconds. (Round your answers to three decimal places.) ?max =0.27 ?'(5) =1.6 is wrong
Figure 2 shows a ballistic pendulum consists of 5 kg slender rod and 10 kg solid cylinder. The pendulum is originally at rest, θ = 0ᵒ. A 10 g bullet with velocity v is fired to the solid cylinder.a) Determine the location of ?̅ of the center of mass G of the pendulum. Then, calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.b) Initially, it is estimated that the value of v is 622 m/s. Analyze the angular velocity of the cylinder just after the bullet embedded. Also, determine the expected value of maximum angle θ will swing when the pendulum stops. Then, evaluate the expected amount of energy lost during impact in percentage.c) However, it is observed that the cylinder swing upward to a maximum angle of θ = 22ᵒ. Determine the actual initial value of v. Then, analyze the linear impulse imparted by the bullet on the cylinder. Also, evaluate the force on the rod immediately just after the impact.
A pendulum consists of a rod of mass 2 kg and length 1 m with a solid sphere at one end with mass 0.3 kg and radius 20 cm (see the following figure). If the pendulum is released from rest at an angle of 30 , what is the angular velocity at the lowest point?
A potters wheel rotates with an angular acceleration = 4at3 3bt2, where t is the time in seconds, a and b areconstants, and has units of radians per second squared. Theinitial position of the wheel is 0 and the initial angular velocityis 0. Obtain expressions for the angular velocity and the angular displacement of the wheel as a function of time.
the wide of a rectangular is measured to be 60.95+/-0.01mm while the length is 221.92+/- 0.01mm. the mass of the bar is measured to be 690.3+/- 0.1g. calculate the moment of inertia of the bad (with uncertainty) using equation I=1/12M(M^2+L^2)
this is a calculus problem A 27-centimeter pendulum moves according to the equation ? = 0.30 cos(3t), where ? is the angular displacement from the vertical in radians and t is the time in seconds. Determine the maximum angular displacement ?max and the rate of change of ? when t = 4 seconds. (Round your answers to three decimal places.) Theta Max=______________________ Theta Derivative' (4)=________________________
mass m = 0.16 kg attached to a thread of length L = 1.5 m The pendulum is released from the horizontal position A as shown in the figure. The angular velocity of the mass as it passes through point B, where β = 120 °, is:
Suppose you start an antique car by exerting a force of 350 N on its crank for 0.16 s.Randomized Variablesf = 350 Nt = 0.16 sd = 0.22 m What angular momentum is given to the engine if the handle of the crank is 0.22 m from the pivot and the force is exerted to create maximum torque the entire time?
If I pull a pendulum to the right, what direction does the angular position vector point? Also, what direction does the torque point?
What is the couple moment if F= 10 N and d= 2m.