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Chapter 15: Q112P (page 443)

In Fig. 15-64, ademolition ball swings from the end of a crane. The length of the swinging segment of cable is 17. (a) Find the period of the swinging, assuming that the system can be treated as a simple pendulum. (b) Does the period depend on the ball鈥檚 mass?

Short Answer

Expert verified
  1. Period of swinging by assuming the system as a simple pendulum is 8.3 S.
  2. The period of a simple pendulum does not depend on the ball鈥檚 mass.

Step by step solution

01

The given data

  • Mass of the ball,m=2500kg.
  • Length of a cable,L=17m..
02

Understanding the concept of SHM

Using the formula of the period of the simple pendulum we can find the period of swinging, and hence, we can determine whether the period of the simple pendulum depends on the ball鈥檚 mass or not.

Formula:

The period of oscillation,T=2蟺Lg

03

(a) Calculation of period

From equation (i), we can find the period of oscillations of the ball as:

T=2蟺179.8m/s2=8.3s

Therefore, the period of swinging by assuming the system as a simple pendulum is

04

(b) Checking whether the period of the ball depends on the ball’s mass or not

From equation (i) of the period of a body鈥檚 oscillation, it can be clearly seen that the term 鈥減eriod鈥� only depends on the body鈥檚 length and acceleration.

From this relation, we can say that period is independent of the mass because this equation does not contain term .

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Most popular questions from this chapter

Question: In Figure, a stick of lengthL = 1.85oscillates as a physical pendulum.

  1. What value of distance x between the stick鈥檚 center of mass and its pivot pointOgives the least period?
  2. What is that least period?

A particle executes linear SHM with frequency 0.25Hz about the point x=0. Att=0, it has displacement x=0.37cm and zero velocity. For the motion, determine the (a) period, (b) angular frequency, (c) amplitude, (d) displacement x(t), (e) velocity v(t), (f) maximum speed, (g) magnitude of the maximum acceleration, (h) displacement at t=3.0s, and (i) speed att=30s.

A 55.0 gblock oscillates in SHM on the end of a spring with k = 1500 N/maccording to x=xmcos(蝇t+蠒). How long does the block take to move from positionto +0.800xm(a) position +0.600xmand (b) position+0.800xm?

Figure 15-34 shows block 1 of mass 0.200kgsliding to the right over a frictionless elevated surface at a speed of. The block undergoes an elastic collision with stationary block, which is attached to a spring of spring constant1208.5N/m. (Assume that the spring does not affect the collision.) After the collision, block2 oscillates in SHM with a period of 0.140s, and block 1 slides off the opposite end of the elevated surface, landing a distance from the base of that surface after falling height h=4.90m. What is the value role="math" localid="1655106415375" ofd?

The center of oscillation of a physical pendulum has this interesting property: If an impulse (assumed horizontal and in the plane of oscillation) acts at the center of oscillation, no oscillations are felt at the point of support. Baseball players (and players of many other sports) know that unless the ball hits the bat at this point (called the 鈥渟weet spot鈥� by athletes), the oscillations due to the impact will sting their hands. To prove this property, let the stick in Fig. simulate a baseball bat. Suppose that a horizontal force F鈫�(due to impact with the ball) acts toward the right at P, the center of oscillation. The batter is assumed to hold the bat at O, the pivot point of the stick. (a) What acceleration does the point O undergo as a result ofF鈫�? (b) What angular acceleration is produced by F鈫�about the center of mass of the stick? (c) As a result of the angular acceleration in (b), what linear acceleration does point O undergo? (d) Considering the magnitudes and directions of the accelerations in (a) and (c), convince yourself that P is indeed the 鈥渟weet spot.
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