Circular Motion Problems

Transcription

1 CHAPER 0 Circular otion CHAPER Circular otion Problems Numerical Problems.. Calculate the angular speed of the second hand, minute hand and hour hand of a clock. Gien: S = 60 s, = 60 min, H = hr o find: S =?, =?, H =? For second hand = S = 60 s 3.4 S S For minute hand = = 60 min = s For hour hand = H = hr = s 3.4 H H 4 Angular speed of second hand = Angular speed of minute hand =.746 x 0-3, Angular speed of hour hand =.455 x What is the angular elocity of the minute hand of a clock? What is the angular displacement of the minute hand in 0 minutes? If the minute hand is 5 cm long, what is the linear elocity of its tip? Gien: = 60 min, t = 0 min = 0 60 s, r = 5 cm = m o find: =?, =?, =? For minute hand = = 60 min = s Now, t 3 t rad rad 3 - JUS EPOWERING YOU

2 CHAPER 0 Circular otion Problems Solutions Now, Angular speed of minute hand =.746 x 0-3, Angular displacement of minute hand =.095 rad Linear speed of tip of minute hand = 8.73 x 0-5 m/s. 3. What is angular displacement of the minute hand of a clock in 5 minutes? Gien: = 60 min = 60 x 60 s, t = 5 min. 3 5 r m/s o find: =? t 5 t rad rad 6 Now, 3 5 r Angular displacement of minute hand =.68 rad 4. What is the angular elocity of a second hand of a clock? If the second hand is 0 cm long find linear elocity of its tip. Gien: S = 60 s, r = 0 cm = m o find: =?, =?. S For second hand = S = 60 s 3.4 S Now, S Angular speed of second hand = 0.05, r m/s Linear speed of tip of second hand =.05 x 0 - m/s =.05 cm/s - JUS EPOWERING YOU

3 CHAPER 0 Circular otion 3 5. he second hand of a watch is.5 cm long. Find the linear speed of a point on the second hand at a distance of 0.5 cm from the tip. Gien: S = 60 s, r =.5 cm cm = cm = 0 - m o find: =?. For second hand = S = 60 s 3.4 S S Linear speed of the point on second hand =.05 x 0-3 m/s = 0.05 cm/s 6. he extremity of the hour hand of a clock moes /0th as fast as the minute hand. What is the length of the hour hand if the minute hand is 0 cm long? 3 r m/s Gien: H, r 0 = 0 cm, = 60 min, H = hr o find: r H =?. H 0 rh H r 0 rh r 0 H H rh r = 6 cm Length of hour hand = 6 cm 7. Calculate angular elocity of the earth due to its spin motion. Gien: For earth = 4 hr = s, o find: =? Angular speed of theearth due to its spin motion is JUS EPOWERING YOU

4 4 CHAPER 0 Circular otion Problems Solutions 8. A turntable rotates at 00 re/min. Calculate its angular speed in and degrees/s. Gien: N = 00 r.p.m. o find: and degrees/s. N degrees/s 3 Angular speed = 0.47, Angular speed = 600 degrees/s 9. Propeller blades of an aero plane are m long. When the propeller is rotating at 800 re/min, compute the tangential elocity of the tip of the blade. Also find the tangential elocity at a point on blade midway between tips and axis. Gien: n = r = m, N = 800 r.p.m., o find: =?, =? when r = / = m.. Part - I N r m/s Part - II r m/s Angular speed of tip of blade = 377, Angular speed of point midway = he length of an hour hand of a wrist watch is.5 cm. Find the magnitudes of following w.r.t. tip of the hour hand a) angular elocity b) linear elocity c) angular acceleration d) radial acceleration e) tangential acceleration f) linear acceleration Gien: r =.5 cm =.5 0- cm o find: =?, =?, =?, a CP =?, a =?, a =? w.r.t tip of hour hand For hour hand = H = hr = s Angular elocity H 4 - JUS EPOWERING YOU

5 Linear elocity 4 6 r m/s CHAPER 0 Circular otion 5 Angular acceleartion As tip of hour hand performs uniform ircular motion. = 0. Radial acceleration a r m/s. angential acceleration: As tip of hour hand performs uniform ircular motion. Linear acceleration a a a r CP m/s. Angular speed = Linear speed = Angular acceleration = 0 a = 0. Radial acceleration = m/s. angential acceleration = 0 Linear acceleartion = m/s.. A turntable has a constant angular speed of 45 r.p.m. Express this in rad per second and degrees per second. If the radius of the turn table is 0.5 m, what is the linear speed of a point on the rim? Gien: N = 45 r.p.m., r = 0.5 m o find: and degrees/s. N degrees/s r m/s Angular speed = 4.73 = 70 degrees/s Linear speed on point on rim =.357 m/s. - JUS EPOWERING YOU

6 6 CHAPER 0 Circular otion Problems Solutions. he linear elocity of a point on the rotating disc is 3 times greater than at a point on the at a distance of 8 cm from it. What is the diameter of the disc. Gien: = 3 P, r = r, r P = (r - 8) cm o find: Diameter of the disc =? Let r be the radius of the disc. Angular elocity for both the points is the same. 3 r r P 3r P P 3r P r 3( r 8) r 3r 4 r 4 = diameter Diameter of disc = 4 cm. 3. A disc has a diameter of one metre and rotates about an axis passing through its centre and at right angles to its plane at the rate of 0 re/min. What are the angular and linear speed of a point on the rim and at a point halfway to the centre. Gien: d = m, r = 0.5 m, N = 0 r.p.m., o find: =?, V =? for r = r/ = 0.5 m P P N For point on rim Angular speed =.57 Linear speed r m/s For point on point halfway Angular speed =.57 Linear speed r m/s Angular speed of point on rim =.57, Linear peed of point on rim = 6.84 m/s, Angular speed of point on halfway =.57m/s Linear speed of point on halfway3.4 m/s. - JUS EPOWERING YOU

7 CHAPER 0 Circular otion 7 4. A body rotates in a circular path of radius 0.5 m at 40 r.p.m. Find its angular and linear speeds. If the angular speed changes to 330 r.p.m. in 0 s. Find the angular and linear accelerations. Gien: r = 0.5 m, N = 40 r.p.m.,n = 330 r.p.m., t = 0 s. o find: =?, a =? Initial condition Angular elocity N Linear elocity r m/s Final condition Angular elocity N Linear elocity r m/s Angular acceleration t 0 Linear acceleration a r m/s. Initial angular elocity = 5.4 Initial linear elocity = 6.84 m/s Final angular elocity = Final linear elocity = 8.64 m/s Angular acceleration = 0.94 Linear acceleration = m/s. 5. A disc is rotating in a horizontal plane about a ertical axis passing through its centre at 50 r.p.m. When accelerated its speed increases to 050 r.p.m. in 3 s. What is the angular acceleration caused assuming it to be uniform? What will be the angular elocity of the disc in r.p.m. after more seconds? How many rotations does it makes during this time and what is the angular displacement? Gien: N =50 r.p.m.,n =050 r.p.m., t = 3 s. o find: =?, N =? after sec, =?. No. of reolutions =? N 50 5 N JUS EPOWERING YOU

8 8 CHAPER 0 Circular otion Problems Solutions Part - I: t 3 3 Part - II N = 50 r.p.m., N =?, t = 3 + = 5 s N N 60t t N N N 50 N r.p.m. Part - III t t 35 () (0 )() rad 90 Number of rotations = 45 Angular acceleration =3.4, Angualr elocity after sec = 650 r.p.m. No. of rotations in these sec = he angular acceleration of a body rotating about a gien axis is. hrough what angle does it rotates during the time in which its angular elocity increases from 5 to 5. Gien: =, = 5, = 5,. o find: =?. (5) (5) () rad Angle traced = 00 rad. - JUS EPOWERING YOU

9 CHAPER 0 Circular otion 9 7. A disc rotating about an axis passing through its centre and right angles to its plane has its angular speed reduced 50 r.p.s. to 5 r.p.s. in 5 s. How many reolutions does it make during this time? How much time does it take and how many more reolutions does it make before coming to rest. Gien: n = 50 r.p.s., n = 5 r.p.s., t = 5 s. o find: =?. n n 5 50 Part -I t 5 5 t t 00 (5) ( 0 )(5) rad 375 Number of rotations = 87.5 Part -II t t 0 t t 5 s 0 50 (5) ( 0 )(5) rad Number of rotations = No. of reolutions made in 5 sec =87.5, ime taken by disc before coming to rest = 5 s, No. of rolution made before coming to rest = he angular elocity of a disc rotating in a horizontal plane about a ertical axis passing through its centre increases from 600 r.p.m. to 3000 r.p.m. in 5 s. Find the angular acceleration of the disc assuming it to be constant. What are the initial and final angular elocities of the disc? What is the angular displacement and number of reolutions made by the disc during this time? Find the linear elocity of a point on the rim of the disc if its radius is 0.5 m. Gien: N = 600 r.p.m.,n = 3000 r.p.m., t = 5s, r = 0.5 m o find: =?, No. of reolutions =?, elocity of point on rim =? N JUS EPOWERING YOU

10 0 CHAPER 0 Circular otion Problems Solutions N t t t 0 (5) (6 )(5) rad 300 Number of rotations = 50 r m/s Initial angular elocity = 6.84 Final angular elocity = 34. Angular acceleration =50.7, Angualr displacementc = 300 rad No. of reolutions = 50 Linear elocity of point on rim = 57. m/s. 9. A satellite reoles around the earth in a circular orbit of radius 7000 km. If the period of reolution is hours, calculate its angular speed, linear speed and centripetal acceleration. Gien: r = 7000 km = m, = hr = s = 700 s o find: =?, =?, a cp =? r m/s acp = 6. km/s r (8.780 ) m/s. Angular speed = 8.78 x 0-4, Linear speed 6. km/s, Centripetal acceleration = 5.33 m/s. - JUS EPOWERING YOU

11 CHAPER 0 Circular otion 0. Find the speed at a point on the equator moe as the earth rotates about its axis. ake radius of the earth as 6400 km. Gien: r = 6400 km = m, = 4 hr = s = 700 s o find: =?, =?, a cp =? r r m/s Linear speed of point on equator = m/s. he tangential acceleration of a particle moing in a circular path of radius 5 cm is m/s. Angular elocity of the particle increases from 0 to 0 during this time. Find the duration of time and number of reolutions completed during this time. Gien: a = m/s, r = 5 cm = m, = 0, = 0 o find: t =?, No. of reolutions =?, a r a 40, r 50 t t 4 0 t 0 0 t 0.5 s 40 t t 0(0.5) (40)(0.5) rad Number of rotations = =0.6 ime interal = 0.5 s, No. of rotations = JUS EPOWERING YOU