Physics, 11.10.2019 03:30 amazeking3545

An object of mass m swings in a horizontal circle on a string of length l that tilts downward at angle θ. find an expression for the angular velocity ω in terms of g, l and angle θ

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Answer from: jada741

The angular velocity of the object moving in the horizontal circle is given as $\boxed{\omega = \sqrt {\frac{g}{{L\sin \theta }}}}$ .

Further Explanation:

When the object moves in the horizontal circle about a fixed axis, then the vertical component of the tension in the wire is balanced by the weight of the object and the horizontal component of the tension in the wire is balanced by the centripetal force acting in the outward direction as shown in the figure below.

Let there be tension developed in the string due to the motion of the block in the horizontal circle and the mass of the block is .

The centripetal force acting on the block due to its motion in the circular path of radius is:

${F_c} = m{\omega ^2}r$

The force balancing equation on the box in the vertical direction is:

$T\sin \theta = mg$ .......(1)

The force balancing equation on the box in horizontal direction is:

$T\cos \theta = m{\omega ^2}r$ ......(2)

Divide equation (1) by (2).

$\dfrac{{T\sin \theta }}{{T\cos \theta }} = \dfrac{{mg}}{{m{\omega ^2}r}}$

Substitute $L\cos \theta$ for in above expression.

$\begin{aligned}\tan \theta &= \frac{g}{{{\omega ^2}L\cos \theta }}\\\omega &= \sqrt {\frac{g}{{L\cos \theta \tan \theta }}} \\&=\sqrt {\frac{g}{{L\sin \theta }}}\\\end{aligned}$

Thus, the angular velocity of the object moving in the horizontal circle is given as $\boxed{\omega = \sqrt {\frac{g}{{L\sin \theta }}}}$ .

Learn More:

1. If forces acting on an object are unbalanced, the object could experience a change in , direction, or both

2. A toy train rolls around a horizontal 1.0-m-diameter track. The coefficient of rolling friction is 0.10

3. A 30.0-kg box is being pulled across a carpeted floor by a horizontal force

Answer Details:

Grade: College

Subject: Physics

Chapter: Uniform Circular Motion

Keywords: Angular velocity, horizontal circle, tilts downward, angle theta, acceleration due to gravity, weight, centripetal force, outward, and force balancing.

Answer from: Quest

im sorry i dont know

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

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