In recent years it has become possible to "weigh" individual DNA molecules by measuring the influence of their mass on a nanoscale cantilever (oscillator), as shown in the figure below. The figure shows a thin rectangular cantilever etched out of silicon. The cantilever has a mass M, where M = 3.7 X 10-16 kg. If pulled and released, the end of the cantilever vibrates, moving up and down like a diving board after a jump. When bathed with DNA molecules that bind to the cantilever's surface, the addition of the DNA mass causes a small but measurable change in its oscillation frequency. For an experiment, the nano-cantilever was initially vibrating at exactly 12 MHz. After attachment of a DNA molecule, its frequency of oscillation decreased by 50 Hz. Determine the mass of the DNA molecule in Daltons? For example, 1 kg = 6.022 x 1026 Daltons. Hint: The nano-cantilever can be modeled as block of mass mc oscillating on a spring with constant k, where m = 1/3M. The spring constant k does not change when mass is added to the spring. So the value of the spring constant is the same before and after DNA is added. Therefore use this fact to calculate k. In other words, calculate k of the nano-cantilever system before the mass of the DNA was added to it. 4000 nm 400 nm Thickness 100 nm

Figure P10.16 shows the drive train of a bicycle that has wheels 67.3 cm in diameter and pedal cranks 17.5 cm long. The cyclist pedals at a steady cadence of 76.0 rev/min. The chain engages with a front sprocket 15.2 cm in diameter and a rear sprocket 7.00 cm in diameter. Calculate (a) the speed of a link of the chain relative to the bicycle frame, (b) the angular speed of the bicycle wheels, and (c) the speed of the bicycle relative to the road. (d) What pieces of data, if any, are not necessary for the calculations? Figure P10.16

A grandfather clock has a pendulum length of 0.7 m and mass bob of 0.4 kg. A mass of 2 kg falls 0.8 m in seven days to keep the amplitude (from equilibrium) of the pendulum oscillation steady at 0.03 rad. What is the Q of the system?

Consider the simplified single-piston engine in Figure CQ15.13. Assuming the wheel rotates with constant angular speed, explain why the piston rod oscillates in simple harmonic motion.

Consider the simplified single-piston engine in Figure CQ12.13. Assuming the wheel rotates with constant angular speed, explain why the piston rod oscillates in simple harmonic motion. Figure CQ12.13

Review. A string is wound around a uniform disk of radius R and mass M. The disk is released from rest with the string vertical and its top end tied to a fixed bar (Fig. P10.78). Show that (a) the tension in the string is one third of the weight of the disk, (b) the magnitude of the acceleration of the center of mass is 2g/3, and (c) the speed of the center of mass is (4gh/3)1/2 after the disk has descended through distance h. (d) Verify your answer to part (c) using the energy approach. Figure P10.78

A watch balance wheel (Fig. P15.25) has a period of oscillation of 0.250 s. The wheel is constructed so that its mass of 20.0 g is concentrated around a rim of radius 0.500 cm. What are (a) the wheels moment of inertia and (b) the torsion constant of the attached spring? Figure P15.23

A 50.0-g object connected to a spring with a force constant of 35.0 N/m oscillates with an amplitude of 4.00 cm on a frictionless, horizontal surface. Find (a) the total energy of the system and (b) the speed of the object when its position is 1.00 cm. Find (c) the kinetic energy and (d) the potential energy when its position is 3.00 cm.

The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

A lightweight spring with spring constant k = 225 N/m is attached to a block of mass m1 = 4.50 kg on a frictionless, horizontal table. The blockspring system is initially in the equilibrium configuration. A second block of mass m2 = 3.00 kg is then pushed against the first block, compressing the spring by x = 15.0 cm as in Figure P16.77A. When the force on the second block is removed, the spring pushes both blocks to the right. The block m2 loses contact with the springblock 1 system when the blocks reach the equilibrium configuration of the spring (Fig. P16.77B). a. What is the subsequent speed of block 2? b. Compare the speed of block 1 when it again passes through the equilibrium position with the speed of block 2 found in part (a). 77. (a) The energy of the system initially is entirely potential energy. E0=U0=12kymax2=12(225N/m)(0.150m)2=2.53J At the equilibrium position, the total energy is the total kinetic energy of both blocks: 12(m1+m2)v2=12(4.50kg+3.00kg)v2=(3.75kg)v2=2.53J Therefore, the speed of each block is v=2.53J3.75kg=0.822m/s (b) Once the second block loses contact, the first block is moving at the speed found in part (a) at the equilibrium position. The energy 01 this spring-block 1 system is conserved, so when it returns to the equilibrium position, it will be traveling at the same speed in the opposite direction, or v=0.822m/s. FIGURE P16.77