# A child abuses an antique record player to further his knowledge of physics. The record turntable can be treated as a uniform disk IMR2) with mass 0.75 kg and radius 0.16 m that spins freely. First, the child wraps a long string several times around the turntable's circumference. He then pulls horizontally on the string with a constant linear acceleration of 1.2 m/s2, causing the turntable to begin rotating from rest. Assuming that the string doesn't slip, what angular speed will the turntable have once the child has pulled the end of the string a distance of 2.0 m? Next, the child drops a handful of spaghetti onto the turntable (the spaghetti sticks to the turntable). The child then observes that the rotation rate of the turntable is now 5.0 rad/s slower. Find the moment of inertia of the spaghetti about its rotation axis. If you didn't answer the previous part, use a value of 12 rad/s for the initial angular velocity of the turntable (so it ends up at 7.0 rad/s) To stop the spinning mess, the child could press his thumb against the turntable near the rotation axis or near the outer edge of the turntable. Which choice would result in the child needing to do the LEAST amount of work to stop the turntable, or is it a tie? Justify your answer with a conceptual argument or a calculation

(b)1.83s

(c)0.03 kgm2

(d)At the outer edge

Explanation:

The moments of inertia of the record table can be calculated as:

(a) If he is pulling with a constant linear acceleration of a= 1.2 m/s2, then the constant angular acceleration is

(b)The time it takes for the child to pull a distance of s = 2m given a = 1.2 m/s2

Then the angular speed the turntable would have achieved by that time is

(c) By the law of conservation in angular momentum:

where I1 is the initial moment of the turn table before spaghetti drop, and I2 is after.

(d) For the same force, the child could generate different amount of torque, depending on where he's pressing his thumb. If it's near the the rotational axis, the moment arm is very small, or not at all, making the torque small. If it's at the edge, then the moment arm is large, making greater torque, so less work.

## Related Questions

A small resort is situated on an island that lies exactly 3 miles from pp, the nearest point to the island along a perfectly straight shoreline. 10 miles down the shoreline from pp is the closest source of fresh water. if it costs 1.5 times as much money to lay pipe in the water as it does on land, how far down the shoreline from pp should the pipe from the island reach land in order to minimize the total construction costs?

Suppose the pipe goes underwater directly from the resort to point P, which is x miles away from the point on the shoreline that's closest to the resort. By Pythagoras, the distance from the resort to P is: sqrt(x^2 + 3^2) = sqrt(x^2 + 9) Suppose that it costs 1 unit of money per mile to lay pipe on land, so therefore it costs 1.5 units of money to lay pipe underwater. So the cost of the pipe above is: 1.5*sqrt(x^2 + 9) So the distance from P to the fresh water source is 10 - x, so the cost of laying that pipe is 10 - x. Total cost of pipe: C = 1.5*sqrt(x^2 + 9) + 10 - x dC/dx = ((1.5x) / sqrt(x^2 + 9)) - 1 To minimize C, set the derivative to zero: ((1.5x) / sqrt(x^2 + 9)) - 1 = 0 (1.5x) / sqrt(x^2 + 9) = 1 1.5x = sqrt(x^2 + 9) (1.5x)^2 = x^2 + 9 2.25x^2 = x^2 + 9 2.25x^2 - x^2 = 9 1.25x^2 = 9 x^2 = 9 / 1.25 x = sqrt(9/1.25) x = 2.683

2.683

Explanation:

Which of the following is not a force causing plate motion? A. Gravity
B. Ridge push
C. Slab pull
D. Basal drag

D IS WRONG

As per plate tectonics theory therer are three forces that cause plate motion: basal drag, ridge push and slab pull.

Basal drag is caused by convection currents in the mantle.

Ridge push is the name for the force due to the fact that the mid-ocen ridges are higher (more elevated) than the seafloor, which causes that the surrounding rocks are pulled down and away from the ridge pushing the plates.

Slab pull refers to the fact than when a slab sinks it pulls on the rest of the plate.

Name a common product produced by blow molding.

Parts made from blow molding are plastic, hollow, and thin-walled, such as bottles and containers that are available in a variety of shapes and sizes. Small products may include bottles for water, liquid soap, shampoo, motor oil, and milk, while larger containers include plastic drums, tubs, and storage tanks.

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = t3 − 12t2 + 36t (a) Find the velocity at time t. v(t) = 3t2−24t+36 (b) What is the velocity after 5 s? v(5) = ft/s (c) When is the particle at rest? (Enter your answer as a comma-separated list.) t = 6,2 (d) When is the particle moving in the positive direction? (Enter your answer in interval notation.) t = (2,6) (e) Find the total distance traveled during the first 8 s. ft

a) is the velocity of the particle at any time t.

b)

c) The particle is at rest at time t=2 seconds.

d) At time t=2 seconds the particle posses positive velocity being at positive direction.

e)

Explanation:

Given:

The function of displacement dependent on time, .......(1)

a)

Now as we know that velocity is the time derivative of the displacement:

..........................(2)

b)

Now the velocity after 5 seconds:

put t=5 in eq. (2)

c)

When the particle is at rest it has zero velocity.

Now velocity at time t=6 s:

Now velocity at time t=2 s:

The particle is at rest at time t=2 seconds.

d)

Put the value t=2 sec. in eq. (1):

At time t=2 seconds the particle posses positive velocity being at positive direction.

e)

Distance travelled during the first 8 seconds:

put t=8 in eq. (1)

What will the weather most likely be like the day after a cold front?

As a cold front arrives, you are, most likely in the warm sector of the frontal low pressure. You could say that the air will be warmer than after the passing of the front.

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of your rich uncle. The top of the fence is 5.00 m above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of 1.60 m above the ground and at an angle of 56.0o above the horizontal. (a) What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? (b) For the initial velocity calculated in part (a), what horizontal distance be-yond the fence will the rock land on the ground?

Explanation:

a ) Height to be cleared = 5 - 1.6 = 3.4 m

Horizontal distance to be cleared = 5 m .

angle of throw = 56°

here y = 3.4 , x = 5 , θ = 56

equation of trajectory

y = x tanθ - 1/2 g ( x/ucosθ)²

3.4 = 5 tan56 - 1/2 g ( 5/ucos56)²

3.4 = 7.4 - 122.5 / .3125u²

122.5 / .3125u² = 4

u² = 98

u = 9.9 m /s

Range = u² sin 2 x 56 / g

= 9.9 x 9.9 x .927 / 9.8

= 9.27 m

horizontal distance be-yond the fence will the rock land on the ground

= 9.27 - 5

= 4.27 m

a)

b)

Explanation:

Given:

horizontal distance between the fence and the point of throwing,

height of the fence from the ground,

height of projecting the throw above the ground,

angle of projection of throw from the horizontal,

• Let the minimum initial speed of projection of the throw be u meters per second so that it clears the top of the fence.
• Now the effective target height,

The horizontal component of the velocity that remains constant throughout the motion:

Now the time taken to reach the distance of the fence:

use equation of motion,

.................................(1)

Now the time taken to reach the fence height (this height must be attained on the event of descending motion of the rock for the velocity to be minimum).

Maximum Height of the projectile:

........................(4)

Now the height descended form the maximum height to reach the top of the fence:

time taken to descent this height from the top height:

where:

initial vertical velocity at the top point

time of descend

..............................(2)

So we find the time taken by the rock to reach the top of projectile where the vertical velocity is zero:

where:

initial vertical velocity

final vertical velocity

time taken to reach the top height of the projectile

.................................(3)

Now the combined events of vertical and horizontal direction must take at the same time as the projectile is thrown:

So,

Max height:

Now the rock hits down the ground 1.6 meters below the level of throw.

Time taken by the rock to fall the gross height :

Time taken to reach the the top of the fence from the top, using eq. (2):

Time difference between falling from top height and the time taken to reach the top of fence:

b)

Now the horizontal distance covered in this time:

is the horizontal distance covered after crossing the fence.

Apply a force of 50N to the left describe the motion of the box

When a force is applied to the box , this will cause an acceleration to the box.
(force =mass×acceleration)

So the box has a constant acceleration and a changing velocity.

True or false a microchip lets computers process infomation very quickly