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FrozenT [24]
3 years ago
9

A top is a toy that is made to spin on its pointed end by pulling on a string wrapped around the body of the top. The string has

a length of 51 cm and is wrapped around the top at a place where its radius is 1.8 cm. The thickness of the string is negligible. The top is initially at rest. Someone pulls the free end of the string, thereby unwinding it and giving the top an angular acceleration of 10 rad/s2. What is the final angular velocity of the top when the string is completely unwound
Physics
1 answer:
AlladinOne [14]3 years ago
3 0

Given Information:

Angular displacement = θ = 51 cm = 0.51  m

Radius = 1.8 cm = 0.018 m

Initial angular velocity = ω₁ = 0 m/s

Angular acceleration = α = 10 rad/s ²

Required Information:

Final angular velocity = ω₂ = ?

Answer:

Final angular velocity = ω₂ = 21.6 rad/s

Explanation:

We know from the equations of kinematics,

ω₂² = ω₁² + 2αθ

Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.

ω₂² = (0)² + 2αθ

ω₂² = 2αθ

ω₂ = √(2αθ)

We know that the relation between angular displacement and arc length is given by

s = rθ

θ = s/r

θ = 0.51/0.018

θ = 23.33 radians

finally, final angular velocity is

ω₂ = √(2αθ)

ω₂ = √(2*10*23.33)

ω₂ = 21.6 rad/s

Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.

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1. An express train, traveling at 36 m/s, is accidentally sidetracked onto a local train track. The express engineer spots a loc
Colt1911 [192]

Answer:

(i) 12 seconds

(ii) 216 meters from the initial position

(iii) 132 meters from the initial position

(iv) No

Explanation:

Speed of express train =36 m/s

Speed of local train =11 m/s

The initial distance between the local train and passenger train =100 m.

Due to the application of breaks, the express train slows at the rare of 3.0 m/s^2.

So, the acceleration of the express train, a=-3 m/s^2.

(i) Let t be the time the express train takes to stop.

From the equation of motion,

v=u+at

where, v: final velocity, u: initial velocity, a: constant acceleration, t: time taken to change the speed from u to v.

In this case, v=0, u=36 m/s, a=-3 m/s^2

So, 0=36+(-3)t

\Rightarrow t= 36/3=12 seconds.

(ii) To compute the distance traveled, s, till the express train stops, using

v^2=u^2+2as

\Rightarrow 0^2=36^2+2(-3)s

\Rightarrow s=\frac{36\times36}{6}

\Rightarrow s=216 meters.

(iii) The local train is moving at a speed of 11 m/s

So, in 12 seconds, the distance, d, traveled by the local train

d= 11x12=132 meters [as distance= speed x time]

(iv) Let 0 be the reference position which is the initial position of the express train.

So, at the initial time, the position of the local train is at 100m.

After 12 seconds:

The position of the express train is at 216 m [using part (ii)]

and the position of the local train is at 100+132=232m  [using part (iii)].

So, the local train is still ahead of the express train, hence the trains didn't collide.

6 0
3 years ago
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