IB DP Physics Topic 2: Mechanics Question Bank SL Paper 3


A wheel of mass 0.25 kg consists of a cylinder mounted on a central shaft. The shaft has a radius of 1.2 cm and the cylinder has a radius of 4.0 cm. The shaft rests on two rails with the cylinder able to spin freely between the rails.

The stationary wheel is released from rest and rolls down a slope with the shaft rolling on the rails without slipping from point A to point B.

The wheel leaves the rails at point B and travels along the flat track to point C. For a short time the wheel slips and a frictional force F exists on the edge of the wheel as shown.

a.i.The moment of inertia of the wheel is 1.3 × 10–4 kg m2. Outline what is meant by the moment of inertia.[1]

a.ii.In moving from point A to point B, the centre of mass of the wheel falls through a vertical distance of 0.36 m. Show that the translational speed of the wheel is about 1 m s–1 after its displacement.[3]
a.iii.Determine the angular velocity of the wheel at B.[1]
b.i.Describe the effect of F on the linear speed of the wheel.[2]
b.iiDescribe the effect of F on the angular speed of the wheel.[2]



an object’s resistance to change in rotational motion


equivalent of mass in rotational equations


[1 mark]


ΔKE + Δrotational KE = ΔGPE


\(\frac{1}{2}\)mv2 + \(\frac{1}{2}\)I\(\frac{{{v^2}}}{{{r^2}}}\) = mgh

\(\frac{1}{2}\) × 0.250 × v2 + \(\frac{1}{2}\) × 1.3 × 10–4 × \(\frac{{{v^2}}}{{1.44 \times {{10}^{ – 4}}}}\) = 0.250 × 9.81 × 0.36

v = 1.2 «m s–1»

[3 marks]


ω «= \(\frac{{1.2}}{{0.012}}\)» = 100 «rad s–1»

[1 mark]


force in direction of motion

so linear speed increases

[2 marks]


force gives rise to anticlockwise/opposing torque on

wheel ✓ so angular speed decreases ✓


[2 marks]



A uniform ladder of weight 50.0 N and length 4.00 m is placed against a frictionless wall making an angle of 60.0° with the ground.

a. Outline why the normal force acting on the ladder at the point of contact with the wall is equal to the frictional force F between the ladder and the ground. [1]

b. Calculate F. [2]

c. The coefficient of friction between the ladder and the ground is 0.400. Determine whether the ladder will slip. [2]



a. «translational equilibrium demands that the» resultant force in the horizontal direction must be zero✔

«hence NW =

Equality of forces is given, look for reason why.


«clockwise moments = anticlockwise moments»

50 × 2cos 60 = NW × 4sin 60 ✔

« N W = F = 50 × 2 cos 60 4 sin 60 »

F = 14.4«N» ✔


maximum friction force = «0.4 × 50N» = 20«N» ✔

14.4 < 20 AND so will not slip ✔

Scroll to Top