Circular Motion. Radians. One revolution is equivalent to which is also equivalent to 2π radians. Therefore we can.

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1 1 Cicula Moion Radians One evoluion is equivalen o which is also equivalen o 2π adians. Theefoe we can say ha 360 = 2π adians, 180 = π adians, 90 = π 2 adians. Hence 1 adian = 360 2π Convesions Rule 1: Convein fom deees o adians you muliply by 2π 360 which is equivalen o π Rule 2: Convein fom adians o deees you muliply by 360 2π 180 which is equivalen o 180 π Example 1 Conve: (a) (b) (c) (d) 45 0 o adians o adians 3π 5 2π 3 adians o deees adians o deees Anula Displacemen (Ѳ) An objec movin alon a cicula pah passes houh an anle Ѳ measued a he cene of he cicle as shown in he diaam below. The anle Ѳ is called he anula displacemen and i is measued in adians. We can calculae his anula displacemen Ѳ by usin: θ = s

2 2 Anula Velociy (ω) When an objec moves alon he ac a a consan speed, we can calculae he anula velociy by usin he fomula: ω = θ whee θ is he anle in adians ω is he anula velociy in adians pe seconds is he ime in seconds Peiod (T) The peiod is he ime aken fo he objec o complee one evoluion (houh 2π adians ). We can calculae he peiod by usin he equaion: T = 2π ω Insananeous Anula Velociy If he anula velociy is no consan, we can wok i ou by usin he fomula below: ω = v We can deive his fomula by fis calculain he linea velociy which is he ae of chane of disance alon he cicumfeence (s) wih ime (). v = s We know ha: θ = s s = θ Hence Bu we ae iven ha v = θ ω = θ eq (i) Theefoe we can ewie eq (i) as v = ω and hence ω = v

3 3 Cenipeal Foce If a body is movin in a cicula pah, a foce mus be acin on i o i will move off in a saih line (accodin o Newon s Fis Law). If he body is also movin a a seady speed, his foce canno have a componen which is he diecion of moion of he body (else he body will incease and decease in speed). The foce heefoe acs pependiculaly o he diecion of he moion of he body and is dieced owads he cene of he cicula pah. This foce is known as he cenipeal foce. Example: If a bick is spun in a cicle on he end of a piece of ope, he ension in he ope povides he cenipeal foce. If he ope beaks, he bick will fly off a a anen. Cenipeal Acceleaion An objec movin in a cicula pah wih a consan speed has a chanin velociy. This occus because he diecion of he velociy chanes even houh is maniude emains he same. Such an objec mus have an acceleaion, which is acin in he same diecion as he foce, owads he cene of he cicle. This acceleaion is known as he cenipeal acceleaion. We can use he fomula below o calculae he cenipeal acceleaion of an objec a = ω 2 If we ae iven he linea speed of he objec we can also he followin equaion a = v2 If we use Newon s Second Law which calculaes he esulan foce by usin he fomula (F = ma), we will e he followin equaions. F = m ω 2 F = m v2

4 4 Deivin ha he cenipeal acceleaion (a = v2 ) Le us conside a paicle movin a a consan speed (v) alon an ac NOP (as shown in he diaam above). The componen of he velociy of he paicle has he same value a P as a N. Theefoe he acceleaion of he x- componen is zeo. We know ha acceleaion = speed ime Hence fo he x-componen, he acceleaion (ax) can be calculaed by: a x = (v cos θ v cos θ) a x = 0 The y componen of acceleaion (ay) can be iven by: a y = v sin θ ( v sin θ) a y = v sin θ + v sin θ a y = 2v sin θ eq i

5 5 We can also calculae he speed of he paicle alon he ac by usin he fomula: speed (v) = ac lenh (NOP) = ac lenh (NOP) speed (v) We can find he ach lenh (OP) by usin s = θ Theefoe he ac lenh (NOP) would be 2s = 2θ Hence we can hen say ha = 2θ speed (v) eq ii If we eaane equaion i o e = 2v sin θ a y and hen equae o equaion ii, we e 2v sin θ a y = 2θ v Hence a y = v2 sin θ θ If θ is vey small hen sin θ θ Then we can say ha a y = v2

6 6 Banked Tack Conside a vehicle of mass (m) k, avellin on a banked cicula ack. The speed of he vehicle is consan. Suppose he ack is inclined a he anle Ѳ as shown in he diaam below. We can make he followin saemens. Saemens: The weih of he vehicle (W) acs veically downwads The Nomal (R) acs pependicula o he ack. The Nomal (R) can be sepaaed ino is veical and hoizonal componens (as shown in he diaam above) The cenipeal foce equied o make he vehicle o aound he bend is povided by he foce exeed on he yes by he oad (he Nomal, R). Since he vehicle of mass (m) is movin wih a consan speed (v) aound he bend of adius (), he cenipeal foce will povide an acceleaion which can be calculaed by usin he equaion: a y = v2 If we use Newon s Second Law (F = ma), we e ha F = mv2 Since he vehicle acquies a hoizonal componen as a esul of bankin we can ewie he equaion o sae R sin θ = mv2 eq i Noe ha hee is no veical acceleaion bu we can make he saemen ha R cos θ = m eq ii

7 7 If we divide equaion i by equaion ii, we e ha R sin θ R cos θ = an θ = mv2 mv 2 m m an θ = mv2 1 m an θ = v2 This equaion can be applied o ailway ains avellin aound a bend on level acks and o aiplanes bankin in ode o make a un (See Muncase Pae 73). Moion in a Veical Cicle If we use a small mass which is headed on wie of lenh () and movin in a cicle in he veical plane, we can calculae he ension in he wie by usin he followin equaion. T m cos θ = mv2 T = mv2 + m cos θ When small mass a BOTTOM: θ = 0 T = mv2 + m When small mass a TOP: θ = 180 T = mv2 m

8 8 The Conical Pendulum Conside a mass (m) is aached o a sin of lenh (L). Suppose he mass avels in a hoizonal cicula pah so ha he adius () and he sin is a anle Ѳ o he veical, as shown in he he diaam below. Noe: Thee ae wo foces acin on he pendulum bob The weih of he bob (W = m) The ension in he sin (F) The hoizonal componen of he ension povides he cenipeal foce and can be calculaed by usin he equaion T sin θ = mv2 eq i Since hee is no veical acceleaion, he veical componen can be calculaed by T cos θ = m eq ii If we divide equaion i by equaion ii, we e ha an θ = v2 eq iii (NB: same as he equaion used in objecs ha ae bankin)

9 9 We know ha v = ω eq iv If we subsiue equaion iv ino equaion iii, we e ha an θ = v2 an θ = (ω)2 an θ = ω2 2 Fom he diaam above we see ha an θ = ω2 = L sin θ Hence an θ = ω2 L sin θ Bu Theefoe an θ = sin θ cos θ sin θ cos θ = ω2 L sin θ Makin ω he subjec, we can wie ha ω 2 = sin θ cos θ L sin θ Bu ω = 2π T Hence ( 2π T ) 2 = Lcos θ

10 10 4π 2 T 2 = Lcos θ T 2 = 4π2 Lcos θ Squae ooin boh sides we e ha T = 4π2 Lcos θ T = 2π Lcos θ

PHYS PRACTICE EXAM 2

PHYS PRACTICE EXAM 2 PHYS 1800 PRACTICE EXAM Pa I Muliple Choice Quesions [ ps each] Diecions: Cicle he one alenaive ha bes complees he saemen o answes he quesion. Unless ohewise saed, assume ideal condiions (no ai esisance,