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Peoples Empowerment Group
ISBM COLLEGE OF ENGINEERING, NANDE, PUNE
DEPARTMENT OF APPLIED SCIENCE
Academic Year 2019-20-
ENGINEERING MECHANICS
Experiment no. 5
Title: To study the rolling motion of a sphere on a surface and also the
trajectory of spinning sphere.
Aim : To study the rolling motion of a sphere on a surface and also the trajectory of spinning
sphere.
Apparatus : A metallic track shaped as an arc of circle which is fixed on to a plate from .The
track allows a sphere kept at the top of track to roll down to either side one of which students an the angle of 37o^ and other a full semicircle.
Theory:
Consider a rolling without sliding a body of radius ‘r’ released from top of the circular track of radius R as shown in fig. If the radius of gyration of the rolling body is K ,it can be shown using kinematics and kinetics of a rigid body that the velocity V of the body after covering angle as shown in fig. The radius of sphere is given by V =
2 gR +( 1 +
r
R )
( 1 −cos ө )
(^1 +^
k 2
r 2 )^
The angle ө, at which the body makes and exits from the track is given by Өe=cos־I[2/ (3+(k^2 r^2 )]…………………..(2) The velocity of exit Vc can be the force obtain by substituting the eq.2 in eq 1
V=
{ [ 2 qR (^) ( 1 + r R )] [ 3 +(^ k 2 − r 2 )^ ] } For a sphere k 2 r 2 =2/5 and if it is relatively small compared to track for r/R < 0.04 ,so r/R can be neglected in eq 2 for a relatively small rolling body. Thus V=(10g/7)R(1-cosθ) Ө = cos-I(
)=54^0
Vc=√(10g/17)R Consider the motion of the sphere after it makes it exist either free or force, from the track. If the exist is make an angle as shown in fig2 .The equation of trajectory is given by Y = (x-Rsinθ) tanθ + g(x-Rsinθ )^2 / (2V^2 COS^2 θ )…………………(3) Where the exist velocity is given by eg. If sphere allow strike and horizontal surface passing through the bottomof the fig. y = R + Rcosθ The trajectory eq thus yields. R(1+cosθ) = (x/Rsinθ) tanθ + g(x/Rsinθ )^2 / 2((10/7)gR(1-cosθ))cos^2 θ……………..(4) Eq.4 can be expressed as quadratic eq in (x/R) as 7(x/R)^2 + sinθ (20cos^2 θ-14)(x/R) + sin^2 (7-20cos θ) = 0……(5) Which on solving gives (x/R) =1. For force exit θ = 37^0 , Eq 5 reduces to 7(x/R)^2 - 6.4898(x/R) - 2.3497 = 0 Which on solving gives (x/R) = 1.
c) Sphere No 2. a) b) c) Sphere No 3. a) b) c) B) Forced Exit: (x/R) Sphere No 1. a) b) c) Sphere No 2. a) b) c) Sphere No 3. a) b) c) Difference in Free Exit:
- (Mean – Analytical)/mean X 100 Difference in Forced Exit:
- (Mean-Analytical)/mean X 100
Result:
Comparison of experimental value of x/R with the analytical value is given in observation table.
Conclusion:
The main reason for the difference in experimentally and analytical value of ‘x’ is due to the force applies to the sphere and it exactly passes and spin after leaving the track though ti is treated as particle during that motion. Manual recording of x/R precise and sensitive job and hence any lapse on close observation affects the result.
Figure :-
