Home » KINEMATICS
Category Archives: KINEMATICS
All students of class XI are to copy down the questions and solve it in the home work copy and solve it to submit on or before 22 Sept 2014.
- What is the angular velocity of a second hand and minute hand of a clock?
- A body of mass 0.4 kg is whirled in a horizontal circle of radius 2 m with a constant speed of 10 ms . Calculate its (i) angular speed (ii) frequency of revolution (iii) time period and (iv) centripetal acceleration.
- A circular wheel of 0.50 m radius is moving with a speed of10 ms. Find the angular speed.
- Assuming that the moon completes one revolution in a circular orbit around the earth in 27.3 days, calculate the acceleration of the moon towards the earth. The radius of the circular orbit can betaken as 3.85 x 10 km.
- The angular velocity of a particle moving along a circle of radius 560 cm is increased in 5 minutes from 100 revolutions per minute to 400 revolutions per minute. Find angular acceleration and (ii) linear acceleration.
- Calculate the linear acceleration of a particle moving in a circle of radius 0.4 m at the instant when its angular velocity is 2 rad /s and its angular acceleration is 5 rad/s2
- A threaded rod with 12 turns per cm and diameter 1.18 cm is mounted horizontally. A bar with a threaded hole to match the rod is screwed onto the rod. The bar spins at the rate of 216 rpm. How long will it take for the bar to move 1.50 cm along the rod.
Is it possible to create a rocket with a variable blast diameter to achieve different levels of torque and power similar to a geared transmission?
I suppose the best analogy I have is the difference in power between a garden hose normally and when you plug it with your thumb, creating more pressure.
Would it be possible to have a rocket launch with maximum diameter, and slowly close up like a camera shutter to gain speed once in the air? Perhaps this shudder would be angled like a funnel to direct the blast easier. All the rockets I see have a fixed diameter and it just makes me curious.
Experts in the field are requested to answer via comments
How it is possible to double the momentum of a body by increasing the kinetic energy four times?at the time what happen to mass?
When we say, KE is increased, we mean to say by increasing speed alone unless otherwise specified.
Here mass remains constant.
Please note that the question says, KE of a body increased 4 times; which suggests that the body is the same and KE is increased by increasing the speed alone.
I just want to know how to easily solve problems related to kinematics and friction. I don’t know why i always face problem in these two topics a lot……
Please help and tell some tricks to solve these both…….
In a circus show a clown who is sitting on a post with height,H drops an apple to the ground.At the same time another clown throw an orange vertically upward from ground with speed 11 m/s.If both fruits pass each other after 0.8s, (a) determine the height (b) where do the apple and orange meet? (c) what is the speed of each fruit when they meet each other?
[Asked nur zainun syafiqah ayob]
If a person jumps out of a plane, and a few seconds later, another person jumps, why is it that the difference in their velocities stays the same? Shouldn’t it decrease because eventually, the first diver will reach maximum velocity, but the other diver will be slower at first, but then he reaches max. velocity as well…I’m not wording it the best way, but any help would be greatly appreciated :]
[Posted by Iris]
Both are examples for free fall. The velocity at any instant is given by v = gt
If for the first one, time = t and for the second who jumps after n seconds, time=t-n
Therefore, velocity of the first one at any instant, v1=gt
for second one, v2=g(t-n)
The difference in velocities,
v1-v2 = gn, which shows that the difference in velocity remains constant.
This is because, both are moving with the same acceleration and their velocities are increasing at the same rate
“A ball is projected vertically upwards with an initial velocity of 28m/s and is observed to pass a point 30m above the projection point,at what 2times does the ball pass these points?”
(Vimal Raj Answered)
ans : 1.356 sec and 1.444 sec .
from v=u+at we get t=2.8 . (time taken to reach the max height )
fm v2-u2=2aS we get S=39.2 m (max height reached)
observer at 30 m .
so , in btw distance btw observer and ball (at the top) =9.2 m
at top ball is at rest and time taken to reach the observer can be calculaed from S=ut + 1/2 at2 ,we get t=1.356s .
and next time can be calculated by 2.8 – 1.356 =1.444 sec .
(The answer has not been scrutinized. If any error, teachers and visitors can post as comment)