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Date	November 2018	Marks available	1	Reference code	18N.3.SL.TZ0.6
Level	Standard level	Paper	Paper 3	Time zone	0 - no time zone
Command term	Calculate	Question number	6	Adapted from	N/A

Question

A uniform rod of weight 36.0 N and length 5.00 m rests horizontally. The rod is pivoted at its left-hand end and is supported at a distance of 4.00 m from the frictionless pivot.

The support is suddenly removed and the rod begins to rotate clockwise about the pivot point. The moment of inertia of the rod about the pivot point is 30.6 kg m².

Calculate the force the support exerts on the rod.

[2]

Calculate, in rad s^–2, the initial angular acceleration $\alpha$ of the rod.

[2]

b.i.

After time t the rod makes an angle θ with the horizontal. Outline why the equation $\theta = \frac{1}{2}\alpha {t^2}$ cannot be used to find the time it takes θ to become $\frac{\pi }{2}$ (that is for the rod to become vertical for the first time).

[2]

b.ii.

At the instant the rod becomes vertical show that the angular speed is ω = 2.43 rad s^–1.

[3]

c.i.

At the instant the rod becomes vertical calculate the angular momentum of the rod.

[1]

c.ii.

Markscheme

taking torques about the pivot R × 4.00 = 36.0 × 2.5 ✔

R = 22.5 «N» ✔

36.0 × 2.50 = 30.6 × $\alpha$ ✔

$\alpha$ = 2.94 «rad s^–2» ✔

b.i.

the equation can be applied only when the angular acceleration is constant ✔

any reasonable argument that explains torque is not constant, giving non constant acceleration ✔

b.ii.

«from conservation of energy» Change in GPE = Change in rotational KE ✔

$W\frac{L}{2} = \frac{1}{2}I{\omega ^2}$ ✔

$\omega = \sqrt {\frac{{36.0 \times 5.00}}{{30.6}}}$ ✔

«ω = 2.4254 rad s^–1»

c.i.

L = 30.6 × 2.43 = 74.4 «J s» ✔

c.ii.

Examiners report

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

Syllabus sections

Option B: Engineering physics » Option B: Engineering physics (Core topics) » B.1 – Rigid bodies and rotational dynamics

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