Ask question

Ask question

All categories

3 years ago

7

A group of particles of total mass 35 kg has a total kinetic energy of 313 J. The kinetic energy relative to the center of mass

is 81 J. What is the speed of the center of mass?

1 answer:

lutik1710 [3]3 years ago

5 0

Answer:

The speed of center of mass is 3.64 m/s.

Explanation:

Given that,

Total mass of the group of particles, m = 35 kg

Total kinetic energy of group of particles, $K_T= 313 J$

The kinetic energy relative to the center of mass is, K = 81 J

We need to find the speed of center of mass. We know that the total kinetic energy is equal to the sum of rotational kinetic energy and the rotational kinetic energy.

$K_T=E+K\\\\E=K_T-K$

Since, $E=\dfrac{1}{2}mv_c^2$ , $v_c$ is the speed of center of mass

$\dfrac{1}{2}mv_c^2=K_T-K\\\\v_c=\sqrt{\dfrac{2(K_T-K)}{m}} \\\\v_c=\sqrt{\dfrac{2(313-81)}{35}}\\\\v_c=3.64\ m/s$

So, the speed of center of mass is 3.64 m/s.

You might be interested in

A compact disc (CD) stores music in a coded pattern of tiny pits 10⁻⁷m deep. The pits are arranged in a track that spirals outwa

Serggg [28]

a) Angular speed of the innermost part: 50 rad/s, outermost part: 21.6 rad/s

b) The length of the track would be 5,550 m

c) The average angular acceleration is $-6.4\cdot 10^{-3}rad/s$

Explanation:

a)

For an object in uniform circular motion, the relationship between angular speed and linear speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circle

Here the linear speed of the track is constant:

v = 1.25 m/s

For the innermost part of the disk,

r = 25.0 mm = 0.025 m

So the angular speed is

$\omega_i = \frac{v}{r_i}=\frac{1.25}{0.025}=50 rad/s$

For the outermost part,

r = 58.0 mm = 0.058 m

So the angular speed is

$\omega_o = \frac{v}{r_o}=\frac{1.25}{0.058}=21.6 rad/s$

b)

The maximum playing time of the disk is

$t=74.0 min \cdot (60 s/min)=4440 s$

If the track was stretched out in a straight line, the motion of the track would be a uniform motion (since it is moving at constant speed), so we can use the equation

$v=dt$

where

v is the linear speed

d is the length coverted in a straight line

t is the time

Substituting v = 1.25 m/s and solving for d,

$d=vt=(1.25)(4440)=5550 m$

c)

The angular acceleration is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular speed

$\omega_i$ is the initial angular speed

t is the time

A CD is read from the innermost part to the outermost part, so we have:

$\omega_i = 50 rad/s$ (at the innermost part)

$\omega_f = 21.6 rad/s$ (at the outermost part)

t = 74.0 min = 4440 s is the time

Substituting, the average angular acceleration is

$\alpha = \frac{21.6-50.0}{4440}=-6.4\cdot 10^{-3}rad/s$

where the negative sign means the CD is decelerating.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

6 0

2 years ago

PLEASE HELPPPPPPPPPPPPP

sertanlavr [38]

Answer: sliding

Explanation:

A snow skier slowing to a stop after skiing down a mountain is an example of sliding friction

8 0

2 years ago

Read 2 more answers

A bike travels at 20mph for 5 hours, then 10 mph for 3 hours. What is the average speed?

d1i1m1o1n [39]

Answer:

10 mph

Explanation:

5 0

2 years ago

Nimfa-mama [501]

Answer:

The answer is Dependent Variable

5 0

2 years ago

A very small object with mass 8.30×10-9 kg and positive charge 6.90×10-9 C is projected directly toward a very large insulating

Rainbow [258]

Answer:

$41.4496148484\ m/s$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

$\sigma$ = Surface charge density = $5.9\times 10^{-8}\ C/m^2$

$\Delta x$ = 0.57-0.26

q = Charge = $6.9\times 10^{-9}\ C$

m = Mass of object = $8.3\times 10^{-9}\ kg$

Electric field due to a sheet is given by

$E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m$

Electric field is given by

$E=\dfrac{V}{d}$

Voltage is given by

$V=E\Delta x$

Kinetic energy is given by

$K=qV$

$\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s$

The initial speed of the object is $41.4496148484\ m/s$

7 0

3 years ago