In Figure 4.27 four particles form a square of edge length

A skater rotates with her arms crossed at an angular speed of 8.0 rad/s and she has a moment of inertia of 100 kg/m2. She extend

solong [7]

Answer:

When her hands extends, her momen of inertia is $4.28\ kg-m^2$ .

Explanation:

Given that,

Initial angular speed, $\omega_i=8\ rad/s$

Initial moment of inertia, $I_1=100\ kg-m^2$

Final angular speed, $\omega_f=7\ rad/s$

Initially, a skater rotates with her arms crossed and finally she extends her arms. The momentum remains conserved. Using the conservation of momentum as :

$I_1\omega_1=I_2\omega_2$

$I_2$ is final moment of inertia

$I_2=\dfrac{I_1\omega_1}{\omega_2}\\\\I_2=\dfrac{100\times 8}{7}\\\\I_2=114.28\ kg-m^2$

So, when her hands extends, her momen of inertia is $4.28\ kg-m^2$ . Hence, this is the required solution.

7 0

3 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

$L$ = length of the meter stick = 1 m

$m$ = mass of the meter stick

$w$ = angular speed of the meter stick as it hits the floor

$v$ = speed of the other end of the stick

we know that, linear speed and angular speed are related as

$v = r w\\w = \frac{v}{r}$

$h$ = height of center of mass of meter stick above the floor = $\frac{L}{2} = \frac{1}{2} = 0.5 m$

$I$ = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

$I = \frac{mL^{2} }{3}$

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

$(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}$