Question

A disk rolls along a flat surface at a constant speed of 10m/s. Diameter is 0.5m. At a particular instant P, at the edge of the disk, is 45 degree from the horizontal at the centre. what is the magnitude acceleration of the point of contact of the disk?

Answer 1

Data given -

Velocity = 10 m/s

Diameter = 0.5 m. i.e Radius of a disk is 0.25m

Rolling motion is a combination of rotation and translation.

First of all, we will calculate Angular Velocity of a disk.

Angular velocity is the rate of change of angular displacement with regards to time for an object rotating about a fixed axis at a constant speed.

Angular velocity is given as ω = v / R

where,

ω = Angular velocity,

v = Linear velocity i.e 10 m/s

R = Radius of disk i.e 0.25 m

Substituting the above values, we get,

ω = v / R

= 10 / 0.25

ω = 40 Rad/ s.

Now, magnitude acceleration of the point of contact of the disk is given by the formula.

a = Rω2

Where,

a = magnitude acceleration in m/s2

R = Radius of disk i.e 0.25 m

ω = Angular velocity i.e 40 Rad/ s.

Substituting the above values, we get,

a = Rω2

a = 0.25 x (40 x 40)

a = 400 m/s2  

So, magnitude acceleration of the point of contact of the disk is 400 m/s2