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2 years ago

7

Let theta denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the Bob is m, the

n the tension in string is

2 answers:

posledela2 years ago

8 0

The tension has to hold the part of the weight in the direction of the string:

T = mg*cos(theta)

Theta=0, whole weight, theta=90, T=0, if the pendulum is horizontal, the string will be loose! Yeah

Harrizon [31]2 years ago

6 0

Answer:

So the tension in the string T =mg cosθ

Explanation:

Now given that

mass of bob = m

Angle from vertical plane =θ

Lets T is the tension in the string

From the diagram

Weight of bob have two component ,first one(mg cosθ) balance the tension in the string and other one(mg sinθ) make the bob to move in the plane.

So the tension in the string T =mg cosθ

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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$

5 0

3 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

5 0

2 years ago

3. a) Your body is made up of several simple machines that help you move. Identify three

Alja [10]

Answer:

Explained below

Explanation:

1) The human arm: This is a type of simple machine called "Lever". In this type of machine, the elbow acts as the fulcrum, the palm serves as the load because that's where we place the load we want to carry. While the inner part of the arm which is the inner part of the elbow represents the effort because that is the joint we mover when making use of our arms.

2) Pulleys: An example of this in the human body is the knee cap where the direction of an applied force is changed. Thus means as it is in motion, it alters the direction for which the quadriceps tendon pulls on the tibia.

3) wheel and axle: An example of this in the human body is the lateral rotation of the shoulder joint medial. The humerus which is the bone between the shoulder and elbow will act as the axle while the rotator will be the will because when it is rotated a little bit, the humerus will move along with it.

8 0

2 years ago

how much energy is required to move a 4.00-microcoulomb charge through a potential difference of 36.0 volts?

musickatia [10]

Answer:

All you gotta do is search it up. ask the same question you asked on here, type it into google, it'll teach you a lot more than your actual teachers.

Explanation:

4 0

3 years ago

The higher an object is raised compared to the ground

zmey [24]

Answer:

B. The more potential energy it has.

Explanation:

I majored in Physics

3 0

2 years ago