Ask question

Ask question

All categories

2 years ago

7

A turntable of radius R1 is turned by a circular rubberroller of radius R2 in contact with it at their outeredges. What is the r

atio of their angular velocities,ω1 / ω2 ?

1 answer:

Gwar [14]2 years ago

5 0

Answer:

The ratio is $\frac{w_1}{w_2} = \frac{R_2}{R_1}$

Explanation:

From the question we are told that

The first radius is $R_1$

The second radius is $R_2$

Generally the angular speed of the turntable is mathematically represented as

$w_1 = \frac{ v_k }{R_1 }$

Generally the angular speed of the rubber roller is mathematically represented as

$w_2 = \frac{ v_k }{R_2 }$

Where $v_k$ is the velocity of both turntable and rubber roller

So

$\frac{w_1}{w_2} = \frac{\frac{v_k}{R_1} }{\frac{v_k}{R_2} }$

$\frac{w_1}{w_2} = \frac{R_2}{R_1}$

You might be interested in

A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball

liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$1.45=\frac{u^2\sin 90}{9.8}$

u=3.77 m/s

Conserving Energy

$E_{bottom}=E_{initial\ point\ at\ cliff}$

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

$\frac{mv^2}{2}=mgh+\frac{mu^2}{2}$

$v^2=u^2+2gh$

$v=\sqrt{u^2+2gh}$

$v=\sqrt{3.77^2+2\time 9.8\times 0.64}$

$v=\sqrt{26.75}=5.17 m/s$

5 0

2 years ago

The 1.5-in.-diameter shaft AB is made of a grade of steel with a 42-ksi tensile yield stress. Using the maximum-shearing-stress

PolarNik [594]

Answer:

T = 0.03 Nm.

Explanation:

d = 1.5 in = 0.04 m

r = d/2 = 0.02 m

P = 56 kips = 56 x 6.89 = 386.11 MPa

σ = 42-ksi = 42 x 6.89 = 289.58 MPa

Torque = T =?

<u>Solution:</u>

σ = (P x r) / T

T = (P x r) / σ

T = (386.11 x 0.02) / 289.58

T = 0.03 Nm.

7 0

2 years ago

A skateboarder rides swiftly up the edge of a bowl-shaped surface and leaps into the air. While in the air, the skateboarder fli

Katen [24]

Answer:

Option(a) is the correct answer to the given question .

Explanation:

The main objective of the angular momentum is evaluating however much the rotational movement as well as the angular velocity in the entity does have.The angular momentum is measured in terms of $kgm^{2 }\ / s$ .

In the given question the skateboarder rides quickly up the bottom of a bowl-shaped surface and climb into the air.it means it is rotational movement also it is not touching anything so it is angular momentum.
All the other option is incorrect because it is not follows the given scenario

3 0

2 years ago

How much force is required to drag a 90 lb. box up this "frictionless" inclined plane? 109 lb. 10 lb. 81 lb. 9 lb.

MrRissso [65]

I believe is 10 lb if not it's 9 lb.

5 0

2 years ago

Read 2 more answers

A monkey weighs 6.00 x 102 N and swings from vine to vine. As the monkey grabs a new vine, both vines make an angle of 35.0° wit

zmey [24]

Answer:

$T=366.23\ N$

Explanation:

Given:

mass of monkey, $w=600\ N$

angle of vine from the vertical, $\theta=35^{\circ}$

Now follow the schematic to understand the symmetry and solution via Lami's theorem.

<u>The weight of the monkey will be balanced equally by the tension in both the vines:</u>

Using Lami's Theorem:

$\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}$

$\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }$

$T=366.23\ N$

4 0

2 years ago