a yo-yo is constructed of three disks: two outer disks of mass m, radius r and thickness d, and an inner disk of mass m, radius r and thickness d. the yo-yo is suspended from the ceiling and then released with the string vertical. calculate the tension in the string as the yo-yo falls. note that when the center of the yo-yo moves down a distance y, the yo-yo turns through an angle y/r, which in turn means that the angular speed w is equal to vcm/4

Answers

Answer 1

The tension in the string as the yo-yo falls is given by the equation T = 2mg.

How is the tension in the string related to the mass of the yo-yo?

When the yo-yo falls, it experiences a downward gravitational force equal to the weight of the yo-yo, which is given by mg, where m is the mass of each disk. Since there are two outer disks and one inner disk, the total weight is 2mg.

The tension in the string provides an upward force to counteract the weight of the yo-yo. To keep the yo-yo in equilibrium, the tension in the string must be equal to the weight of the yo-yo. Therefore, the tension in the string is also equal to 2mg.

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Related Questions

A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has a length of 0. 600m and a mass of 4. 50g.

What is the frequency f1 of the string's fundamental mode of vibration?

Express your answer numerically in hertz using three significant figures

Answers

The frequency f₁ of the string's fundamental mode of vibration is approximately 96 Hz, expressed to three significant figures.

The formula used to determine the frequency of a string's fundamental mode of vibration is given by:

f₁ = (1/2L) √(T/μ)

where:

f₁ is the frequency of the string's fundamental mode of vibration

L is the length of the string

T is the tension in the string

μ is the linear mass density of the string

Given values:

L = 0.600 m

T = 765 N

μ = 0.0075 kg/m

By substituting the values into the formula:

f₁ = (1/2L) √(T/μ)

f₁ = (1/2 × 0.600 m) √(765 N/0.0075 kg/m)

f₁ = (0.300 m) √(102000 N/m²)

f₁ = (0.300 m) (319.155)

f₁ = 95.746 Hz ≈ 96 Hz

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the neurons that select a particular motor program are the . lower motor neurons upper motor neurons in the premotor cortex neurons in the basal nuclei neurons in the cerebellum

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Main answer: The neurons that select a particular motor program are the upper motor neurons in the premotor cortex.

The selection and initiation of specific motor programs in the body are primarily controlled by the upper motor neurons located in the premotor cortex. The premotor cortex, which is a region of the frontal lobe in the brain, plays a crucial role in planning and coordinating voluntary movements. These upper motor neurons receive inputs from various areas of the brain, including the primary motor cortex, sensory regions, and the basal ganglia, to generate the appropriate motor commands.

The premotor cortex acts as a hub for integrating sensory information and translating it into motor commands. It receives input from sensory pathways that carry information about the current state of the body and the external environment. This sensory input, along with the information from other brain regions, helps the premotor cortex determine the desired motor program required to accomplish a particular task.

Once the appropriate motor program is selected, the upper motor neurons in the premotor cortex send signals down to the lower motor neurons in the spinal cord and brainstem. These lower motor neurons directly innervate the muscles and execute the motor commands generated by the premotor cortex. They act as the final link between the central nervous system and the muscles, enabling the execution of coordinated movements.

In summary, while several brain regions are involved in motor control, the upper motor neurons in the premotor cortex play a critical role in selecting and initiating specific motor programs. They integrate sensory information and coordinate with other brain regions to generate motor commands, which are then executed by the lower motor neurons. Understanding this hierarchy of motor control is essential for comprehending the complexity of voluntary movements.

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g what form would the general solution xt() have? [ii] if solutions move towards a line defined by vector

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The general solution xt() would have the form of a linear combination of exponential functions. If the solutions move towards a line defined by a vector, the general solution would be a linear combination of exponential functions multiplied by polynomials.

In general, when solving linear homogeneous differential equations with constant coefficients, the general solution can be expressed as a linear combination of exponential functions. Each exponential function corresponds to a root of the characteristic equation.

If the solutions move towards a line defined by a vector, it means that the roots of the characteristic equation are all real and equal to a constant value, which corresponds to the slope of the line. In this case, the general solution would include terms of the form e^(rt), where r is the constant root of the characteristic equation.

To form the complete general solution, additional terms in the form of polynomials need to be included. These polynomials account for the presence of the line defined by the vector. The degree of the polynomials depends on the multiplicity of the root in the characteristic equation.

Overall, the general solution xt() in this scenario would have a combination of exponential functions multiplied by polynomials, where the exponential functions account for the movement towards the line defined by the vector, and the polynomials account for the presence of the line itself.

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a racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 1 6 m/s. the collision takes 5 0 ms. what is the average acceleration (in unit of m/s 2 ) of the ball during the collision with the wall?

Answers

The average acceleration of the racquetball during the collision with the wall is -280 m/s^2.

To find the average acceleration of the racquetball during the collision with the wall, we can use the formula:
Average acceleration = (final velocity - initial velocity) / time

Given that the racquetball strikes the wall with an initial speed of 30 m/s and rebounds with a final speed of 16 m/s, and the collision takes 50 ms (or 0.05 s), we can substitute these values into the formula:
Average acceleration = (16 m/s - 30 m/s) / 0.05 s
Simplifying this equation, we get:

Average acceleration = (-14 m/s) / 0.05 s
Dividing -14 m/s by 0.05 s gives us an average acceleration of -280 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which means the ball is decelerating during the collision.
Therefore, the average acceleration of the racquetball during the collision with the wall is -280 m/s^2.
The average acceleration of the racquetball during the collision with the wall can be found using the formula:

average acceleration = (final velocity - initial velocity) / time. Given that the initial speed is 30 m/s, the final speed is 16 m/s, and the collision takes 50 ms (or 0.05 s), we can substitute these values into the formula. By subtracting the initial velocity from the final velocity and dividing by the time, we find that the average acceleration is -280 m/s^2.

The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, meaning the ball is decelerating during the collision.

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