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 m^{2}. Outline what is meant by the moment of inertia.[1]

^{–1}after its displacement.[3]

*F*on the linear speed of the wheel.[2]

*F*on the angular speed of the wheel.[2]

**▶️Answer/Explanation**

## Markscheme

a.i.

an object’s resistance to change in rotational motion

*OR*

equivalent of mass in rotational equations

*OWTTE*

*[1 mark]*

ΔKE + Δrotational KE = ΔGPE

*OR*

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

\(\frac{1}{2}\) × 0.250 × *v*^{2} + \(\frac{1}{2}\) × 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 ✓

*OWTTE*

*[2 marks]*

## Question

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]

**▶️Answer/Explanation**

Ans:

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

«hence *N*_{W} = *F»*

*Equality of forces is given, look for reason why.*

b.

«clockwise moments = anticlockwise moments»

50 × 2cos 60 = *N*_{W} × 4sin 60 ✔

«${N}_{\text{W}}=F=\frac{50\times 2\mathrm{cos}60}{4\mathrm{sin}60}$»

*F* = 14.4«*N*» ✔

c.

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

14.4 < 20 * AND *so will not slip ✔