<< Chapter < Page Chapter >> Page >

Time interval

Problem : The angular position of a point on a flywheel is given by the relation :

θ (rad) = - 0.025 t 2 + 0.01 t

Find the time (in seconds) when flywheel comes to a stop.

Solution : The speed of the particle is :

ω = θ t = - 0.025 X 2 t + 0.1

When flywheel comes to a standstill, ω = 0,

0 = - 0.025 X 2 t + 0.1

t = 0.1 0.025 = 100 25 = 4 s

Got questions? Get instant answers now!

Problem : The magnitude of deceleration of the motion of a point on a rotating disk is equal to the acceleration due to gravity (10 m / s 2 ). The point is at a linear distance 10 m from the center of the disk. If initial speed is 40 m/s in anti-clockwise direction, then find the time for the point to return to its position.

Solution : The disk first rotates in anti-clockwise direction till its speed becomes zero and then the disk turns back to move in clockwise direction. In order to analyze the motion, we first convert linear quantities to angular quantities as :

ω i = v i r = 40 10 = 4 rad / s

α = a T r = 10 10 = 1 rad / s 2

When the point returns to its initial position, the total displacement is zero. Applying equation of motion for angular displacement, we have :

θ = ω i t + 1 2 α t 2

0 = 4 X t - 1 2 X 1 X t 2

t 2 8 t = 0

t t 8 = 0

t = 0 o r 8 s

The zero time corresponds to initial position. The time of return to initial position, therefore, is 8 seconds.

Got questions? Get instant answers now!

Angular displacement

Problem : A disk initially rotating at 80 rad/s is slowed down with a constant deceleration of magnitude 4 rad / s 2 . What angle (rad) does the disk rotate before coming to rest ?

Solution : Initial and final angular velocities and angular acceleration are given. We can use ω = ω 0 + α t to determine the time disk takes to come to stop. Here,

ω 0 = 80 rad / s , α = 4 rad / s 2

0 = 80 - 4 t t = 20 s

Using equation, θ = ω 0 t + 1 2 α t 2 , we have :

θ = 80 X 20 - 1 2 X 4 X 20 2 = 1600 - 800 = 800 rad

Got questions? Get instant answers now!

Problem : The initial angular velocity of a point (in radian) on a rotating disk is 0.5 rad/s. The disk is subjected to a constant acceleration of 0. 2 rad / s 2 . in the direction opposite to the angular velocity. Determine the angle (in radian) through which the point moves in third second.

Solution : Here, we need to be careful as the point reverses its direction in the third second ! For ω = 0,

ω = ω 0 + α t 0 = 0.5 - 0.2 t t = 0.5 0.2 = 2.5 s

Thus the disk stops at t = 2.5 second. In this question, we are to find the angle (in rad) through which the point moves in third second – not the displacement. The figure here qualitatively depicts the situation. In the third second, the point moves from B to C and then from C to D. The displacement in third second is BOD, whereas the angle moved in the third second is |BOC| + |DOC|. Where,

∠BOC = displacement between 2 and 2.5 seconds.

∠DOC = displacement between 2.5 and 3 seconds.

Angle measurement

The angular velocity at the end of 2 seconds is :

ω = 0.5 - 0.2 x 2 = 0.1 rad / s

BOC = ω t - 1 2 α t 2 BOC = 0.1 X 0.5 - 1 2 X 0.2 X 0.5 2 BOC = 0.05 - 0.1 X 0.25 = 0.05 - 0.025 = 0.025 rad

The angular velocity at the end of 2.5 seconds is zero. Hence,

DOC = 0 - 1 2 x 0.2 x 0.5 2 BOC = - 0.1 x 0.25 = - 0.025 rad

It means that the point, at t = 3 s, actually returns to the position where it was at t = 2 s. The displacement is, thus, zero.

The total angle moved in third second = |0.025| + |-0.025| = 0.05 rad

Got questions? Get instant answers now!

Problem : The angular velocity of a point (in radian) on a rotating disk is given by |t – 2|, where “t” is in seconds. If the point aligns with the reference direction at time t = 0, then find the quadrant in which the point falls after 5 seconds.

Problem : The area under the angular velocity – time plot and time axis is equal to angular displacement. As required, let us generate angular velocity data for first 5 seconds to enable us draw the requisite plot :

--------------------------------- Time (t) Angular velocity (θ)(s) (rad/s) ---------------------------------0 2 1 12 0 3 14 2 5 3---------------------------------

The angular velocity – time plot is as shown in the figure :

Angular velocity – time plot

The displacement is equal to the area of two triangles :

θ = 1 2 X 2 X 2 + 1 2 X 3 X 3 = 6.5 rad

Thus, the point moves 6.5 rad from the reference direction. Now, one revolution is equal to 2π = 2 x 3.14 = 6.28 rad. The particle is, therefore, in the fourth quadrant with respect to the reference direction.

Got questions? Get instant answers now!

Questions & Answers

A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
Samson Reply
water boil at 100 and why
isaac Reply
what is upper limit of speed
Riya Reply
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
Clash Reply
which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58
sajjad
what is the definition of resolution of forces
Atinuke Reply
what is energy?
James Reply
Ability of doing work is called energy energy neither be create nor destryoed but change in one form to an other form
Abdul
motion
Mustapha
highlights of atomic physics
Benjamin
can anyone tell who founded equations of motion !?
Ztechy Reply
n=a+b/T² find the linear express
Donsmart Reply
أوك
عباس
Quiklyyy
Sultan Reply
Moment of inertia of a bar in terms of perpendicular axis theorem
Sultan Reply
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Yara Reply
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
Rama Reply

Get the best Physics for k-12 course in your pocket!





Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Physics for k-12' conversation and receive update notifications?

Ask