Ask question

Ask question

All categories

Alex_Xolod [135]

2 years ago

6

A wheel with rotational inertia 0.04 kg•m2 and radius 0.02 m is turning at the rate of 10 revolutions per second when a friction

al torque is applied to stop it. How much work is done by the torque in stopping the wheel?

1 answer:

Rus_ich [418]2 years ago

7 0

Answer:

-78.96 J

Explanation:

The workdone by the torque in stopping the wheel = rotational kinetic energy change of wheel.

So W = 1/2I(ω₁² - ω₀²) where I = rotational inertia of wheel = 0.04 kgm², r = radius of wheel = 0.02 m, ω₀ = initial rotational speed = 10 rev/s × 2π = 62.83 rad/s, ω₁ = final rotational speed = 0 rad/s (since the wheel stops)

W = 1/2I(ω₁² - ω₀²) = 1/2 0.04 kgm² (0² - (62.83 rad/s)²) = -78.96 J

You might be interested in

A worker pushes a 7 kg shipping box along a roller track. Assume friction is small enough to be ignored because of the rollers.

Bingel [31]

Answer:

a) Fₓ = 23.5 N

b) Net force = Fₓ

Explanation:

An image of the question as described is attached to this solution.

From the image attached, the forces acting on the box include the weight of the box, the normal reaction of the surface on the box, the applied force on the box and the Frictional force opposing the motion of the box (which is negligible and equal to 0)

a) From the diagram, the horizontal component of the force is

Fₓ = 25 cos 20° = 23.49 N = 25 N

b) Again, from the diagram attached, doing a force balance on the box, in the horizontal direction, we obtain

Net force = Fₓ - Frictional force

But frictional force is 0 N

Net force = Fₓ

Hope this Helps!!!

6 0

2 years ago

Daria was swimming in a friend’s pool yesterday, when she saw that a fly had landed in the water about 5 feet away from her. She

jasenka [17]

Answer:

Daria probably suffers from Entomophobia.

5 0

1 year ago

What is the acceleration during the pushing-off phase, in m/27 Express your answer in meters per second squared. A bush baby, an

Sindrei [870]

Answer:

$a = 12.78 g's$

Explanation:

Height reached by the object after push off is given as

$H = 2.3 m$

$v_f^2 - v_i^2 = 2 a s$

now we have

$0 - v^2 = 2(-9.81)(2.3)$

$v = 6.72 m/s$

now we know that this push last for total distance of 0.18 m

so during the push we will have

$v_f^2 - v_i^2 = 2 a d$

$6.72^2 - 0 = 2a(0.18)$

$a = 125.35 m/s^2$

now in terms of g = 9.81 m/s/s we have

$a = \frac{125.35}{9.81} gs$

$a = 12.78 g's$

6 0

1 year ago

A tennis ball is shot vertically upward in an evacuated chamber inside a tower with an initial speed of 20.0 m/s at time t=0s. W

dalvyx [7]

Answer:

at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Explanation:

At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest

So here we will say that at the highest point of the path the speed of the ball comes to zero

now by the force diagram we can say that net force on the ball due to gravity is given by

$F_g = mg$

now the acceleration of ball is given as

$a = \frac{F_g}{m}$

$a = \frac{mg}{m} = g$

so at the highest point of the path the acceleration of ball is same as acceleration due to gravity

5 0

1 year ago

A 55-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a const

nataly862011 [7]

a) $-1.54 m/s^2$

b) 803.4 N

Explanation:

a)

At the point C (top position of the loop), the pilot feel weightless, so the normal reaction exerted by the seat is zero:

N = 0

Therefore, the equation of the forces at position C is:

$mg=m\frac{v^2}{r}$

where the term on the left is the weight of the pilot and the term on the right is the centripetal force, and where:

$g=9.8 m/s^2$ is the acceleration due to gravity

$v$ is the velocity of the jet at the top

$r=1200 m$ is the radius of the loop

Solving for v,

$v_C=\sqrt{gr}=\sqrt{(9.8)(1200)}=108.4 m/s$

So, this is the velocity of the jet at position C.

The velocity at position A (bottom) is

$v_A=550 km/h =152.8 m/s$

The distance covered by the jet is the length of a semi-circumference of radius r, so

$s=\pi r=\pi(1200)=3770 m$

Since the deceleration of the plane is constant, we can find it by using the following suvat equation:

$v_C^2-v_A^2=2as\\a=\frac{v_C^2-v_A^2}{2s}=\frac{108.4^2-152.8^2}{2(3770)}=-1.54 m/s^2$

b)

The force exerted on the pilot by the seat is equal to the normal force.

At point B (half of the loop), we have:

- The normal force exerted by the seat, N, acting towards the center of the loop

- There are no other forces acting towards the center of the loop, so N must be equal to the centripetal force:

$N=m\frac{v_B^2}{r}$ (1)

where $v_B$ is the velocity at position B.

To find the velocity at position B, we notice that the distance covered by the jet between position A and position B is a quarter of a circle:

$s=\frac{\pi r}{2}=\frac{\pi(1200)}{2}=1885 m$

Since we know the deceleration, we can use the suvat equation to find the velocity at point B:

$v_B^2-v_A^2=2as\\v_B=\sqrt{v_A^2+2as}=\sqrt{152.8^2+2(-1.54)(1885)}=132.4 m/s$

Therefore, we can now use eq.(1) to find the normal force exerted by the seat on the pilot at point B:

$N=(55)\frac{(132.4)^2}{1200}=803.4 N$

6 0

2 years ago