Turn on thread page Beta

Q. Mechanics3 - Review exercise2 (circular motion) watch

    • Thread Starter

    Right, I have been having an absolute torrid time with some of these review questions for whatever reason, this question in particular, I always seem to get just off the answer for the last part of the question, qould like to know where I am going wrong, and the method for answering the question.

    23. A smooth sphere, with centre O and radius a, is fixed to the upper surface of a table. A smooth particle P is placed on the surface of the sphere, at a point A, where OA makes an angle Φ with the upward verticle
    (0 < θ <½π ).

    The particle is released from rest. Show that when OP makes an angle θ with the upward verticle, and that the particle is in contact with the sphere,

    a (∂θ/∂t)² = 2g(cosΦ - cosθ )

    Determine the value of θ when the particle leaves the sphere.

    Given that cosΦ = 3/4, show that the particle hits the table at a distance (3/4)√3 a from the verticle diameter

    I can tell you that the value of θ when the particle leaves the sphere:

    cosθ = 2/3 cosΦ

    But the last bit has really got me curfuddled, any help greatly appreciated

    You know that when the particles leaves the surface:
    a²w² = 2ga(3/4 - 1/2) = ga/2
    => v = aw = sqrt[ga/2], which is the linear speed of the particle when it leaves the sphere.

    You also know that the vertical distance of the particle from the table at the last point of contact is a+acos60=1.5a.

    Since the particle leaves at a direction tangent to the sphere, then using some trig you can find that the intial angle of projection is 60. Assuming the particle falls freely under gravity after it leaves the sphere, then use some projectile analysis.

    Vertical motion:
    s = ut - 0.5gt²
    -1.5a = sqrt[ga/2]sin60 t - 0.5gt²
    0.5gt² - 0.5sqrt[1.5ga]t - 1.5a = 0
    t = (0.5sqrt[1.5ga] ± sqrt[(27/8)ga])/2g

    So, the time until the particle strikes the table is:
    (sqrt[1.5ga] + sqrt[13.5ga])/4g

    Horizontal motion:
    s = vt
    = sqrt[0.5ga]cos60 . (sqrt[1.5ga] + sqrt[13.5ga])/4g
    = (sqrt[3]/4) a

    This is the distance from the vertical at the last point of contact. Now we want to find the distance of this vertical from the vertical diameter, which is:
    a sin60 = (sqrt[3]/2) a

    So the total distance is:
    (sqrt[3]/2) a + (sqrt[3]/4) a = (3/4) sqrt[3] a
    • Thread Starter

    Danke Schon, I forgot to put 1.5a as a negative distance, I keep on doing that :p: ...anyway when doing v.t I was struggling with all the wierd square roots, but sorted them out now.

University open days

  • Manchester Metropolitan University
    Postgraduate Open Day Postgraduate
    Wed, 14 Nov '18
  • University of Chester
    Chester campuses Undergraduate
    Wed, 14 Nov '18
  • Anglia Ruskin University
    Ambitious, driven, developing your career & employability? Aspiring in your field, up-skilling after a career break? Then our Postgrad Open Evening is for you. Postgraduate
    Wed, 14 Nov '18
Have you ever experienced bullying?
Useful resources

Make your revision easier


Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here


How to use LaTex

Writing equations the easy way


Best calculators for A level Maths

Tips on which model to get

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.