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Figure 8-56 shows a solid, uniform cylinder of mass 7.00 kg and radius 0.450 m with a light string wrapped around it. A 3.00-N t

AVprozaik [17]

Answer:

a) The cylinder has an angular acceleration of 3.810 radians per square second, b) The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

Explanation:

The external force exerted on string creates a tension force that tries to move the cylinder in translation, but it is opposed by the friction force between cylinder and ground that generates rolling on cylinder. The Free Body Motion on cylinder-string system is presented below as attachment. Given that cylinder is a rigid body in planar motion, two equations of equilibrium for translation and an equation of equilibrium for rotation are needed to represent the system, which are now described:

$\Sigma F_{x} = T + f = M\cdot R\cdot \alpha$

$\Sigma F_{y} = N - M\cdot g = 0$

$\Sigma M_{G} = (T-f)\cdot R = I_{G}\cdot \alpha$

Where:

$T$ - Tension, measured in newtons.

$f$ - Friction force, measured in newtons.

$M$ - Mass of the cylinder, measured in kilograms.

$R$ - Radius of the cylinder, measured in meters.

$\alpha$ - Angular acceleration, measured in radians per square second.

$N$ - Normal force from ground exerted on cylinder, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$I_{G}$ - Moment of inertia of the cylinder with respect to its center of mass, measured in kilogram-square meters.

The moment of inertia of the cylinder is:

$I_{G} = \frac{1}{2}\cdot M\cdot R^{2}$

a) The angular acceleration is determined by solving on first and third equation after eliminating friction force:

$f = M\cdot R \cdot \alpha - T$

$(T-M\cdot R\cdot \alpha+T) \cdot R = I_{G}\cdot \alpha$

$2\cdot T\cdot R = (I_{G} + M\cdot R^{2})\cdot \alpha$

$\alpha = \frac{2\cdot T\cdot R}{I_{G}+M\cdot R^{2}}$

$\alpha = \frac{2\cdot T \cdot R}{\frac{1}{2}\cdot M\cdot R^{2}+M\cdot R^{2} }$

$\alpha = \frac{4\cdot T}{3\cdot M\cdot R}$

If $T = 3\,N$ , $M = 7\,kg$ and $R = 0.45\,m$ , then:

$\alpha = \frac{4\cdot (3\,N)}{(7\,kg)\cdot (0.45\,m)}$

$\alpha = 3.810\,\frac{rad}{s^{2}}$

The cylinder has an angular acceleration of 3.810 radians per square second.

b) The magnitude of the frictional force can be determined with the help of the following expression:

$f = M\cdot R \cdot \alpha - T$

Given that $T = 3\,N$ , $M = 7\,kg$ , $R = 0.45\,m$ and $\alpha = 3.810\,\frac{rad}{s^{2}}$ , the magnitude of the friction force is:

$f = (7\,kg)\cdot (0.45\,m)\cdot \left(3.810\,\frac{rad}{s^{2}} \right)-3\,N$

$f = 9\,N$

The frictional force has a magnitude of 9 newtons and has the same direction of tension force.

5 0

3 years ago

What is the effeciency of the machine that has an input of 120 and an output of 96?

Rama09 [41]

Efficiency = output divided by input
96/120 = 0.8
or can say 80% as well

6 0

3 years ago

Which type of molecule is added to this genetic sequece might create a insertion mutation? A-G-A-T-G-C-A-G-A-G-T-T-A-C-G-G

Jet001 [13]

D. Adding another base pair will re-arrange your DNA sequences and cause an insertion mutation. This will make your codons group of differently, and possibly give you a BAD mutation. However, sometimes the codons still make the same proteins as its supposed to, the mutation will NOT affect you.

Example:

THE BIG FAT CAT ATE THE RAT

Now, if i were to add a letter (X) to this and make the letters group up in three aka the codons:

THE XBI GFA TCA TAT ETH ERA T

As you can you can see, adding a base pair in a DNA insertion will usually have a negative effect, specifically a insertion mutation.

3 0

3 years ago

8. A student does 1,000 J of work when she moves to her dormitory. Her internal energy is decreased by 3,000 J. Determine the he

Lyrx [107]

Answer:

The heat loss during the process = -4000 J

Explanation:

Work done by the student (W) = - 1000 J

Negative sign on the system is due to work done on the system.

Decrease in internal energy (U) = - 3000 J

We know that heat transfer in the system is given by (Q) = U + W

⇒ Q = - 1000 - 3000

⇒ Q = - 4000 J

This is the value of heat transfer during the process And negative sign indicates that heat loss during the process.

6 0

3 years ago

A child at the top of a slide de has a gravitational store of 1800J. What is the child's maximum kinetic store as he slides down

solmaris [256]

The maximum is 1800 J since energy is not created nor destroyed, it is just transformed. In this example, gravitational potential energy is transformed to kinetic energy.

5 0

2 years ago

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