The challenge is to find the instantaneous acceleration of a point on a sphere when it touches the larger sphere it is rolling inside of. Before finding the acceleration of a point on the circumference of the circle, let’s find that of the centre. The radius of the larger circle is 1 metre. The centre of the small circle is not on the circumference though. It is 25cm away because the small circle has radius 25cm. This means the centre of the smaller centre is 100-75 = 25 centimetres away from that of the larger one. The smaller circle completes a rotation whose circumference is 2π*75cm = 150πcm every 4 seconds. It’s velocity is thus 375π millimetres per second sideways. This velocity changes, and it’s vector’s endpoint will naturally also complete a full circular rotation each second. This circle has radius 1.5π m/s and thus circumference 2π*375π mm/s = 75π2 centimetres per second inward. To complete this circle every second, the acceleration must be 75π2 cm/s2. The same rule applies to smaller circle. The acceleration equals 4π2r/s2 inward times the rate of rotation. For the smaller circle, this gives us 1π2 m/s2 toward it’s centre. When the point touches the rim, these vectors will align, and the point will feel their cumulative acceleration. It’s counterintuitive that the point experiences the most force when it stops, but it’s true. Add 1π2 m/s2 inward and 75π2 cm/s2 inward together to get our answer, 175π2 cm/s2 inward.