Ask question

Ask question

All categories

2 years ago

9

Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a

ngular acceleration when it has an angle of 10.0 degrees from the vertical? A typical pencil has an average length of 15.0 cm and an average mass of 10.0g. Calculate the time for the pencil to hit the ground, assuming that it falls from standing perfectly vertical and maintains this angular acceleration.

1 answer:

kherson [118]2 years ago

6 0

Answer:

α = 17 rad / s² , t = 0.4299 s

Explanation:

Let's use Newton's second angular law or torque to find angular acceleration

τ = I α

W r = I α

The weight is applied in the middle of the pencil,

sin 10 = r / (L/2)

r = L/2 sin 10

The pencil can be approximated by a bar that rotates on one end, in this case its moment of inertia is

I = 1/3 M L²

Let's calculate

mg L / 2 sin 10 = (1/3 m L²) α

α f = 3/2 g / L sin 10

α = 3/2 9.8 / 0.150 sin 10

α = 17 rad / s²

If the pencil has a constant acceleration we can use the angular kinematics relationships, as part of the rest wo = 0

θ = w₀ t + ½ α t²

t = RA (2θ / α )

The angle from the vertical to the ground is

θ = π / 2

t = √ (2 π / (2 α ))

t = √ (π / α )

t = √ (π / 17)

t = 0.4299 s

You might be interested in

An object is located 25.0 cm from a convex mirror. The image distance is -50.0 cm. What is the magnification?

Lynna [10]

Answer:

$\boxed{\sf Magnification \ (m) = 2}$

Given:

Object distance (u) = 25.0 cm

Image distance (v) = -50.0 cm

To Find:

Magnification (m)

Explanation:

$\boxed{\bold{\sf Magnification \: (m) = - \frac{Image \: distance \: (v)}{Object \: distance \: (u)}}}$

Substituting values of Image distance(v) & Object distance (u) in the equation:

$\sf \implies m = - \frac{( - 50)}{25}$

-(-50) = 50:

$\sf \implies m = \frac{50}{25}$

$\sf \implies m = \frac{2 \times \cancel{25}}{ \cancel{25}}$

$\sf \implies m = 2$

4 0

2 years ago

if a net horizontal force of 175 N is applied to a bike whos mass is 43 kg what acceleration is produced

Anna [14]

Explanation:

f=175N

m=43kg

a=?

know

f=ma

a=f/m

a=175/43

a=4.06m/s

3 0

2 years ago

A carousel - a horizontal rotating platform - of radius r is initially at rest, and then begins to accelerate constantly until i

OLga [1]

Answer:

α = (ω²)/8π

Explanation:

The angular acceleration(α) of the carousel can be determined by using rotational kinematics:

ω² =ωo² + 2αθ

Let's make α the subject of this equation ;

ω² - ωo² = 2αθ

α = (ω² −ωo²)/2θ

Now, from the question, since initially at rest, thus, ωo = 0

Also,since 2 revolutions, thus, θ = 2 x 2π = 4π since one revolution is 2π

Plugging in the relevant values to get ;

α = (ω²)/2(4π)

α = (ω²)/8π

7 0

2 years ago

1. Two identical bowling balls of mass M and radius R roll side by side at speed v0 along a flat surface. Ball 1 encounters a ra

UNO [17]

Answer:

1/2

Explanation:

We need to make a couple of considerations but basically the problem is solved through the conservation of energy.

I attached a diagram for the two surfaces and begin to make the necessary considerations.

Rough Surface,

We know that force is equal to,

$F_r = mgsin\theta$

$F_r = \mu N$

$F_r = \mu mg cos\theta$

Matching the two equation we have,

$\mu N = \mu mg cos\theta$

$\mu = tan\theta$

Applying energy conservation,

$\frac{1}{2}mv^2_0+\frac{1}{2}I_w^2 = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{2}{5}mR^2\frac{V_0^2}{R^2} = F_r*d+mgh_1$

$\frac{1}{2}mv^2_0+\frac{mv_0^2}{5} = mgsin\theta \frac{h_1sin\theta}+mgh_1$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_1+gh_1$

$h_1 = \frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})$

Frictionless surface

$\frac{1}{2}mv_0^2+\frac{1}{2}I\omega^2 = mgh_2$

$\frac{1}{2}m_v^2+\frac{1}{2}\frac{2}{5}mR^2\frac{v_0^2}{R^2} =mgh_2$

$\frac{v_0^2}{2}+\frac{v_0^2}{5} = gh_2$

$h_2 = \frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})$

Given the description we apply energy conservation taking into account the inertia of a sphere. Then the relation between $h_1$ and $h_2$ is given by

$\frac{h_1}{h_2} = \frac{\frac{1}{2g}(\frac{v_0^2}{2}+\frac{v_0^2}{5})}{\frac{1}{g}(\frac{V_0^2}{2}+\frac{v_0^2}{5})}$

$\frac{h_1}{h_2} = \frac{1}{2}$

8 0

2 years ago

Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio

Nadya [2.5K]

The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

$PE=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

$PE=(8)(1.6)(2)=25.6 J$

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0

2 years ago

Read 2 more answers