What is the conservation of energy?

A flywheel of diameter 1.2 m has a constant angular acceleration of 5.0 rad/s2. the tangential acceleration of a point on its ri

vodka [1.7K]

We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2

5 0

3 years ago

A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an

Mademuasel [1]

Answer:

The angular acceleration is $0.209\ rad/s^2$

Explanation:

Given that,

Angular velocity, $\omega_{i} = 30.0\ rpm$

Angular velocity, $\omega_{f} = 50.0\ rpm$

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

$\alpha=\dfrac{\Delta \omega}{\Delta t}$

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}$

$\alpha=\dfrac{50.0-30.0}{10.0}$

Now, we change the angular velocity in rad/s.

$\omega=20\times\dfrac{2\pi}{60}$

$\omega=2.09\ rad/s$

$\alpha=\dfrac{2.09}{10.0}$

$\alpha=0.209\ rad/s^2$

Hence, The angular acceleration is $0.209\ rad/s^2$

5 0

3 years ago

Read 2 more answers

please solve this for me ,A garden roller is pulled with a force of 200N acting at an angle of 50 degree with the ground level.f

Bingel [31]

Answer:

The force pulling the roller along the ground is 128.55 N

Explanation:

A force of 200 N acting at an angle of 50° with the ground level

This force is pulled a garden roller

We need to find the force pulling the roller along the ground

The force that pulling the roller along the ground is the horizontal

component of the force acting

→ The force acting is 200 N at direction 50° with ground (horizontal)

→ The horizontal component = F cosФ

→ F = 200 N , Ф = 50

→ The horizontal component = 200 cos(50) = 128.55 N

128.55 N is the horizontal component of the force that pulling the

roller along the ground

<em>The force pulling the roller along the ground is 128.55 N</em>

8 0

3 years ago

A car initially traveling at 15.0 m/s accelerates at a constant rate of 4.50 m/s2 over a distance of 45.0 m. How long does it ta

anygoal [31]

Answer:

2.24 seconds

Explanation:

xf = xo + vo t + 1/2 at^2

45 = 0 + 15 t + 1/2 (4.5) t^2

2.25 t^2 + 15t - 45 = 0 Quadratic formula shows t = 2.24 seconds

4 0

2 years ago

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.460 Hz. The pendulum ha

zlopas [31]

Answer:

The moment of inertia is $I =1.0697 \ kg m^2$

Explanation:

From the question we are told that

The frequency is $f = 0.460 \ Hz$

The mass of the pendulum is $m = 2.40 \ kg$

The location of the pivot from the center is $d = 0.380 \ m$

Generally the period of the simple harmonic motion is mathematically represented as

$T = 2 \pi * \sqrt{ \frac{I}{ m * g * d } }$

Where I is the moment of inertia about the pivot point , so making I the subject of the formula it

=> $I = [ \frac{T}{2 \pi } ]^2 * m* g * d$

But the period of this simple harmonic motion can also be represented mathematically as

$T = \frac{1}{f}$

substituting values

$T = \frac{1}{0.460}$

$T = 2.174 \ s$

So

$I = [ \frac{2.174}{2 * 3.142 } ]^2 * 2.40* 9.8 * 0.380$

$I =1.0697 \ kg m^2$

4 0

3 years ago