Physics, 19.03.2020 04:50 demotte79991

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 like a pendulum, passing through its lowest position with angular speed 5.71 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

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Answer from: amyrose90999

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.

Answer from: Quest

Apossible long term health effects of stress would be work

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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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 sid...

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