All categories

3 years ago

when a solid wheel rotates about a fixed axis do all of the points of the wheel have the same tangential speed?

Physics

2 answers:

Gelneren [198K]3 years ago

4 0

Answer:

No - points with different radial distance from the center will experience different tangential speed.

Explanation:

Picture two points on the wheel: point A is very close (say, 1 mm) to the center, point B is further (say, 1m) away from the center. As the wheel makes one whole turn (2pi), both points will follow a circular trajectory, but point A will have traveled (2pi)x(0.001) meters and point B (2pi)x(1) meters in the same time. Point B's tangential velocity is therefore 1000 times greater than that of the point A.

The formula for tangential speed is as follows:

v = (angular velocity in radians/s) x (radius in m)

vitfil [10]3 years ago

3 0

Yes they should all be going the same speed.

Hope this helped :)

You might be interested in

Classify the type of plate boundary where the appalachian mountains formed . how have they changed since their formation

disa [49]

<span>The Appalachian Mountains were formed when colliding tectonic plates folded and upthrust, mainly during the Permian Period and again in the Cretaceous Period. The folds and thrusts were then eroded and carved by wind, streams and glaciers. These erosive processes are ongoing, and the topography of the Appalachian Mountains continue to change. They have changed with the miles of land that are cleared of all vegetation and topsoil. In the 1970's coal miners literally blow away the top of a mountain to get to the coal underneath.</span>

6 0

3 years ago

A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t

julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

$F_k = F_{W,E}$

$kx_1 = mg$

The extension of the spring due to the weight of the object on Moon is a value of $x_2$ , then

$kx_2 = mg_m$

Recall that gravity on the moon is a sixth of Earth's gravity.

$kx_2 = m\frac{g}{6}$

$kx_2 = \frac{1}{6} mg$

$kx_2 = \frac{1}{6} kx_1$

$x_2 = \frac{1}{6} x_1$

We have that the displacement at the earth was $x_1 = 0.3m$ , then

$x_2 = \frac{1}{6} 0.3$

$x_2 = 0.05m$

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0

3 years ago

Crystal kick the 3.0 kg soccer ball horizontally off of the top of a building the building is 6.0 m tall and the ball lands 10.0

sattari [20]

Answer:

gametes

nucleic acids

Explanation:

amino acids

proteins

3 0

3 years ago

Which example describes a nonrenewable resource?

ioda

Answer: Examples of nonrenewable resources include crude oil, natural gas, coal, and uranium. These are all resources that are processed into products that can be used commercially. For example, the fossil fuel industry extracts crude oil from the ground and converts it to gasoline.

8 0

2 years ago

A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s

TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

$vf^{2} =vo^{2}+2*g*h\\$

We clear the vo (initial speed)

$vo=\sqrt{vf^{2}-2*g*h }$

$v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}$

vo=5.87m/s

7 0

3 years ago