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A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing

VARVARA [1.3K]

Explanation:

(a) The given data is as follows.

Length of the rod, L = 0.83 m

Mass of the rod, m = 110 g = 0.11 (as 1 kg = 1000 g)

At the lowest point, angular speed of the rod ( $\omega$ ) = 5.71 rad/s

First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.

I = $I_{CM} + mh^{2}$

= $\frac{1}{12}mL^{2} + m(\frac{L}{2})$

= $\frac{1}{12} \times 0.11 \times (0.83)^{2} + 0.11 \times \frac{0.83}{2}$

= 0.00631 + 0.415

= 0.42131 $kg m^{2}$

Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.

K = $\frac{1}{2}I \omega^{2}$

= $\frac{1}{2} \times 0.42131 kg m^{2} \times (5.71)^{2}$

= 6.86 J

Hence, kinetic energy of the rod at its lowest point is 6.86 J.

(b) According to the conservation of total mechanical energy of the rod, we have

$K_{i} + U_{i} = K_{f} + U_{f}$

$K_{i} = U_{f} - U_{i}$

or, mgh = K = 6.86 J

Therefore, h = $\frac{6.86}{mg}$

= $\frac{0.63}{0.11 \times 9.8}$

= 0.584 m

Hence, the center of mass rises 0.584 m far above that position.

6 0

3 years ago

Read 2 more answers

Two closed containers look the same but when is packed with LED and the other with a few feathers how could you determine which

Rudik [331]

You would want to push both containers, first one and then the other,
with exactly the same strength of push.

The one that takes off slower (with less acceleration from the same
net force) is the one with more mass.

3 0

3 years ago

Read 2 more answers

A 1.40-kg block is on a frictionless, 30 ∘ inclined plane. The block is attached to a spring (k = 40.0 N/m ) that is fixed to a

Oliga [24]

Answer:

the mass drop by 6.5cm before coming to rest.

Explanation:

Given that:

the mass of the block M= 1.40 kg

angle of inclination θ = 30°

spring constant K = 40.0 N/m

mass of the suspended block m = 60.0 g = 0.06 kg

initial downward speed = 1.40 m/s

The objective is to determine how far does it drop before coming to rest?

Let assume it drops at y from a certain point in the vertical direction;

Then :

Workdone by gravity on the mass of the block is:

$w_1g = 1.4*9.81*y*sin30$ = - 6.867y

Workdone by gravity on the mass of the suspended block is:

$w_2g = 0.06*9.81$ = 0.5886

The workdone by the spring = -1/2ky²

= -0.5 × 40y²

= -20 y²

The net workdone is = -20 y² - 6.867y + 0.5886y

According to the work energy theorem

Net work done = Δ K.E

-20 y² - 6.867y + 0.5886 = 1/2 × 0.06 × 1.4²

-20 y² - 6.867y + 0.5886 = 0.5 × 0.1176

-20 y² - 6.867y + 0.5886 = 0.0588

-20 y² - 6.867y + 0.5886 - 0.0588

-20 y² - 6.867y + 0.5298 = 0

multiplying through by (-)

20 y² + 6.867y - 0.5298 = 0

Using the quadratic formula:

$\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$

where;

a = 20 ; b = 6.867 c= - 0.5298

$\dfrac{-(6.867) \pm \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$

$= \dfrac{-(6.867) + \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)} \ \ \ OR \ \ \ \dfrac{-(6.867) - \sqrt{(6.867)^2-(4*20*-0.5298)}}{2*(20)}$ = 0.0649 OR −0.408