Explanation Given: The mass of the rod is 500 g . The length of the rod is 200 m . The radius of the bowl is 200 m . Initially the rod is released from the rest. Draw the free body diagram of the rod and bowl as shown in Figure (1). Write the formula for mass moment of inertia ( I ) of the rod. I = 1 12 m l 2 (I) Here, m is the mass of the rod and l is the length of the rod. Write the formula for kinetic energy ( T ) . T = 1 2 I ω 2 + 1 2 m v G When, v G = ω r . Substitute ω r for v G . T = 1 2 I ω 2 + 1 2 m ( ω r ) (II) Here, I is the mass moment of inertia, v G is the velocity of the rod’s mass center , r is the radial length of the mass center of the rod from the datum and ω is the angular velocity. Refer Figure (1). Write the formula for gravitational potential energy. Here, the gravitational potential energy is equal to total potential energy ( V g = V ) . V = m g y (III) Here, m is the mass of the rod, g acceleration due to gravity, y is the distance from the datum where the mass acting on it. Write the formula for conservation of energy. T 1 + V 1 = T 2 + V 2 (IV) Here, T 1 , T 2 are the initial and final kinetic energy and V 1 , V 2 are the initial and final potential energy. Conclusion: Refer Figure (1a). Calculate the angle ( θ ) . Consider the right angled triangle A E C and take sin θ . sin θ = A E C A Substitute 200 mm for C A and 100 mm for A E . sin θ = 100 mm 200 mm θ = sin − 1 0.5 θ = 30 ° Refer Figure (1a). Calculate the vertical height ( h ) of the mass center of the rod to the datum at initial. Consider the right angled triangle A E D and take sin θ . sin 30 ° = h 100 mm h = 100 × sin 30 ° h = 50 mm × 1 m 1000 mm h = 0.05 m Refer Figure (1b). Calculate the radial length of the mass center of the rod to the datum. Consider the triangle A C E and use Pythagoras theorem. r = C E = ( 200 mm × 1 m 1000 mm ) 2 − ( 100 mm × 1 m 1000 mm ) 2 = 0.2 2 − 0.1 2 = 0.1732 m Calculate the vertical height ( E D ) of the mass center of the rod to the datum at final. E D = C D − C E = 200 mm × 1 m 1000 mm − 0.1732 m = 0.2 m − 0

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

Videos

Question

Chapter 18.5, Problem 61P

To determine

The angular velocity of the rod when it swings downward and becomes horizontal.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 18.5, Problem 60P

Chapter 18.5, Problem 62P

Students have asked these similar questions

100 As a spring is heated, its spring constant decreases. Suppose the spring is heated and then cooled so that the spring constant at time t is k(t) = t sin + N/m. If the mass-spring system has mass m = 2 kg and a damping constant b = 1 N-sec/m with initial conditions x(0) = 6 m and x'(0) = -5 m/sec and it is subjected to the harmonic external force f (t) = 100 cos 3t N. Find at least the first four nonzero terms in a power series expansion about t = 0, i.e. Maclaurin series expansion, for the displacement: • Analytically (hand calculations) Creating Simulink Model Plot solutions for first two, three and four non-zero terms as well as the Simulink solution on the same graph for the first 15 sec. The graph must be fully formatted by code.

Two springs and two masses are attached in a straight vertical line as shown in Figure Q3. The system is set in motion by holding the mass m₂ at its equilibrium position and pushing the mass m₁ downwards of its equilibrium position a distance 2 m and then releasing both masses. if m₁ = m² = 1 kg, k₁ = 3 N/m and k₂ = 2 N/m. (y₁ = 0) www k₁ = 3 Jm₁ = 1 k2=2 www (Net change in spring length =32-31) (y₂ = 0) m₂ = 1 32 32 System in static equilibrium System in motion Figure Q3 - Coupled mass-spring system Determine the equations of motion y₁ (t) and y₂(t) for the two masses m₁ and m₂ respectively: Analytically (hand calculations) Using MATLAB Numerical Functions (ode45) Creating Simulink Model Produce an animation of the system for all solutions for the first minute.

Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from tank A into tank B at a rate of 3 L/min and from B into A at a rate of 1 L/min (see Figure Q1). The liquid inside each tank is kept well stirred. A brine solution with a concentration of 0.2 kg/L of salt flows into tank A at a rate of 6 L/min. The diluted solution flows out of the system from tank A at 4 L/min and from tank B at 2 L/min. If, initially, tank A contains pure water and tank B contains 20 kg of salt. A 6 L/min 0.2 kg/L x(t) 100 L 4 L/min x(0) = 0 kg 3 L/min 1 L/min B y(t) 100 L y(0) = 20 kg 2 L/min Figure Q1 - Mixing problem for interconnected tanks Determine the mass of salt in each tank at time t≥ 0: Analytically (hand calculations) Using MATLAB Numerical Functions (ode45) Creating Simulink Model Plot all solutions on the same graph for the first 15 min. The graph must be fully formatted by code.

Chapter 18 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

Ch. 18.4 - Determine the kinetic energy of the 100-kg object.Ch. 18.4 - The 80-kg wheel has a radius of gyration about its...Ch. 18.4 - The uniform 50-lb slender rod is subjected to a...Ch. 18.4 - The uniform 50-kg slender rod is at rest m the...Ch. 18.4 - The 50-kg wheel is subjected to a force of 50 N....Ch. 18.4 - If the uniform 30-kg slender rod starts from rest...Ch. 18.4 - The 20-kg wheel has a radius of gyration about its...Ch. 18.4 - At a given instant the body of mass m has an...Ch. 18.4 - The wheel is made from a 5-kg thin ring and two...Ch. 18.4 - The wheel is made from a 5-kg thin ring and two...

Ch. 18.4 - A force of P = 60 N is applied to the cable, which...Ch. 18.4 - A force of P = 20 N is applied to the cable, which...Ch. 18.4 - A force of P = 20 N is applied to the cable, which...Ch. 18.4 - The double pulley consists of two parts that are...Ch. 18.4 - The double pulley cons1sts of two parts that are...Ch. 18.4 - Prob. 9P Ch. 18.4 - The spool has a mass of 40 kg and a radius of...Ch. 18.4 - Prob. 11P Ch. 18.4 - Determine the velocity of the 50-kg cylinder after...Ch. 18.4 - The 10-kg uniform slender rod is suspended at rest...Ch. 18.4 - Prob. 14P Ch. 18.4 - The pendulum consists of a 10-kg uniform disk and...Ch. 18.4 - A motor supplies a constant torque M = 6 kNm to...Ch. 18.4 - The center O of the thin ring of mass m is given...Ch. 18.4 - Prob. 18P Ch. 18.4 - Prob. 19P Ch. 18.4 - If P = 200 N and the 15-kg uniform slender rod...Ch. 18.4 - A yo-yo has a weight of 0.3 lb and a radius of...Ch. 18.4 - Prob. 22P Ch. 18.4 - Prob. 23P Ch. 18.4 - Prob. 24P Ch. 18.4 - The 30-kg disk is originally at rest, and the...Ch. 18.4 - Two wheels of negligible weight are mounted at...Ch. 18.4 - Prob. 27P Ch. 18.4 - The 10-kg rod AB is pin connected at A and...Ch. 18.4 - The 10-lb sphere starts from rest at = 0 and...Ch. 18.4 - Motor M exerts a constant force of P = 750 Non the...Ch. 18.4 - Prob. 31P Ch. 18.4 - The linkage consists of two 6-kg rods AB and CD...Ch. 18.4 - The two 2-kg gears A and B are attached to the...Ch. 18.4 - The linkage consists of two 8-lb rods AB and CD...Ch. 18.4 - The linkage consists of two 8-lb rods AB and CD...Ch. 18.5 - F187. If the 30-kg disk is released from rest when...Ch. 18.5 - The 50-kg reel has a radius of gyration about its...Ch. 18.5 - The 60-kg rod OA is released from rest when = 0....Ch. 18.5 - Prob. 10FP Ch. 18.5 - The 30-kg rod is released from rest when = 45....Ch. 18.5 - Prob. 12FP Ch. 18.5 - Prob. 36P Ch. 18.5 - Prob. 37P Ch. 18.5 - An automobile tire has a mass of 7 kg and radius...Ch. 18.5 - The spool has a mass of 20 kg and a radius of...Ch. 18.5 - The spool has o mass of 20 kg and a radius of...Ch. 18.5 - A uniform ladder having a weight of 30 lb is...Ch. 18.5 - Prob. 44P Ch. 18.5 - The 12-kg slender rod is attached to a spring,...Ch. 18.5 - Prob. 46P Ch. 18.5 - The 40-kg wheel has a radius of gyration about its...Ch. 18.5 - The assembly consists of two 10-kg bars which are...Ch. 18.5 - The assembly consists of two 10-kg bars which are...Ch. 18.5 - The compound disk pulley consists of a hub and...Ch. 18.5 - Prob. 51P Ch. 18.5 - Prob. 52P Ch. 18.5 - The two 12-kg slender rods are pin connected and...Ch. 18.5 - If the 250-lb block is released from rest when the...Ch. 18.5 - The slender 15-kg bar is initially at rest and...Ch. 18.5 - If the chain is released from rest from the...Ch. 18.5 - Prob. 57P Ch. 18.5 - Prob. 58P Ch. 18.5 - The slender 6-kg bar AB is horizontal and at rest...Ch. 18.5 - Prob. 60P Ch. 18.5 - Prob. 61P Ch. 18.5 - The 50-lb wheel has a radius of gyration about its...Ch. 18.5 - The system consists of 60-lb and 20-lb blocks A...Ch. 18.5 - The door is made from one piece, whose ends move...Ch. 18.5 - Prob. 65P Ch. 18.5 - Prob. 66P Ch. 18.5 - Prob. 67P Ch. 18.5 - The system consists of a 30-kg disk A, 12-kg...Ch. 18.5 - The pendulum of the Charpy impact machine has a...Ch. 18.5 - Prob. 2RP Ch. 18.5 - The drum has a mass of 50 kg and a radius of...Ch. 18.5 - The spool has a mass of 60 Kg and a radius of...Ch. 18.5 - Prob. 5RP Ch. 18.5 - At the Instant shown, the 50-lb bar rotates...Ch. 18.5 - Prob. 7RP Ch. 18.5 - Prob. 8RP

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The angular velocity of the rod when it swings downward and becomes horizontal.

Solution Summary: The author explains the angular velocity of the rod when it swings downward and becomes horizontal, and the formula for mass moment of inertia and kinetic energy.

Concept explainers

Videos

The angular velocity of the rod when it swings downward and becomes horizontal.

Want to see the full answer?

Chapter 18 Solutions