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Over [174]
3 years ago
13

"a uniform thin rod of length 0.962 m is hung from a horizontal nail passing through a small hole in the rod located 0.048 m fro

m the rod's end. when the rod is set swinging about the nail at small amplitude, what is the period of oscillation?"
Physics
1 answer:
exis [7]3 years ago
5 0
<span>1.57 seconds. The rod hanging from the nail constructs a physical pendulum. The period of such a pendulum follows the formula T = 2*pi*sqrt(L/g) where T = time L = length of pendulum g = local gravitational acceleration So the problem becomes one of determining L. It's tempting to consider L to be the distance between the center of mass and the pivot, but that isn't the right value. The correct value is the distance between the pivot and the center of percussion. So let's determine what that is. We can treat the uniform thin rod as an uniform beam and for an uniform beam the distance between the center of mass and the center of percussion is expressed as b = L^2/(12A) where b = distance between center of mass and center of percussion L = length of beam A = distance between pivot and center of mass Since the rod is uniform, the CoM will be midway from either end, or 0.962 m / 2 = 0.481 m from the end. The pivot will therefore be 0.481 m - 0.048 m = 0.433 m from the CoM Now let's calculate the distance the CoP will be from the CoM: b = L^2/(12A) b = (0.962 m)^2/(12 * 0.433 m) b = (0.925444 m^2)/(5.196 m) b = 0.178107005 m With the distance between the CoM and CoP known, we can now calculate the effective length of the pendulum. So: 0.433 m + 0.178107005 m = 0.611107005 m And finally, with the effective length known, let's calculate the period. T = 2*pi*sqrt(L/g) T = 2*pi*sqrt((0.611107005 m)/(9.8 m/s^2)) T = 2*pi*sqrt(0.062357858 s^2) T = 2*pi*0.249715554 s T = 1.569009097 s Rounding to 3 significant figures gives 1.57 seconds. Let's check if this result is sane. Looking up "Seconds Pendulum", I get a length of 0.994 meters which is longer than the length of 0.611 meters calculated. But upon looking closer at the "Seconds Pendulum", you'll realize that it's period is actually 2 seconds, or 1 second per swing. So the length of the calculated pendulum is sane.</span>
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In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m a
matrenka [14]
Assuming you are looking for the acceleration a:

1.m_1a = T_1 -m_1g
2.m_2a = m_2g - T_2
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3.\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}
where I = \frac{1}{2} mR^2 and a = \alpha R.

Combining the three equations:
T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }
6 0
3 years ago
Is visible light a form of<br> radiation? Explain.
natali 33 [55]

Answer:

Yes,in fact visible 'light' is a form of radiation, which can be defined as an energy that travels in the form of electromagnetic waves. It can also be described as a flow of particle-like 'wave-packets', called photons, that travel constantly at the speed of light (about 300 000 kilometres per second).

Explanation:

6 0
3 years ago
how long would it take for a resultant upward force of 100N to increase the speed of 50kg object from 100m/s to 150m/s​
ValentinkaMS [17]

Answer:

25 sec

Explanation:

F = ma

100 = 50 a

a = 2 m/s^2

vf = vo + at

150 = 100 +  2 t

50 = 2 t

t = 25 sec

7 0
2 years ago
An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

4 0
3 years ago
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2

\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0
3 years ago
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